ATOMIC STRUCTURE
MCQs
Q.1 Splitting of spectral lines when atoms are subjected to strong electric field is called
(a) Zeeman effect (b) Stark effect
(c) Photoelectric effect (d) Compton effect
Q.2 The velocity of photon is
(a) independent of its wavelength
(b) depends on its wavelength
(c) equal to square of its amplitude
(d) depends on its source
Q.3 The nature of positive rays depend on
(a) the nature of electrode
(b) the nature of discharge tube
(c) the nature of residual gas
(d) all of the above
Q.4 The wave number of the light emitted by a certain source is 2 x 106 m. The wavelength of this light is
(a) 500 nm (b) 500 m
(c) 200 nm (d) 5 x 10–1 m
Q.5 Rutherford ’s model of atom failed because
(a) the atom did not have a nucleus and electrons
(b) it did not account for the attraction between protons and neutrons
(c) it did account for the stability of the atom
(d) there is actually no space between the nucleus and the electrons
Q.6 Bohr’s model of atom is contradicted by
(a) Planck’s quantum theory
(b) Pauli exclusion principle
(c) Heisenberg uncertainty principle
(d) All of the above
Q.7 Quantum number value for 2p orbitals are
(a) n = 2, l = 1 (b) n = 1, l = 2
(c) n = 1, l = 0 (d) n = 2, l = 0
Q.8 In the ground state of an atom, the electron is present
(a) in the nucleus (b) in the second shell
(c) nearest to the nucleus (d) farthest from the nucleus
Q.9 When the 6d orbital is complete the entering electron goes into
(a) 7f (b) 7s
(c) 7p (d) 7d
Q.10 Orbitals having same energy are called
(a) hybrid orbitals (b) valence orbitals
(c) degenerate orbitals (d) d–orbitals
Q.11 The e/m value for the positive rays is maximum for
(a) hydrogen (b) helium
(c) nitrogen (d) oxygen
Q.12 Neutron was discovered by Chadwick in
(a) 1935 (b) 1930
(c) 1932 (d) 1934
Q.13 The velocity of photon is
(a) equal to square of its amplitude
(b) independent of its wavelength
(c) Equal to its wave number
(d) equal to the velocity of light
Q.14 Quantum number values for 3p orbitals are
(a) n = 0, l = 3 (b) n = 3, l = 1
(c) n = 2, l = 1 (d) n = 1, l = 3
Q.15 The radius of first orbit of hydrogen atom
(a) 0.329 Ao (b) 0.429 Ao
(c) 0.529 Ao (d) 0.229 Ao
Q.16 All atoms are principally composed of few fundamental particles which are in number
(a) 2 (b) 3
(c) 4 (d) 5
Q.17 Which scientist gave the name of electron to the cathode rays
(a) Planck (b) Einstein
(c) Stoney (d) Bohr
Q.18 The divisibility of atom was showed by
(a) Stoney (b) J.J. Thomson
(c) Millikan (d) Rutherford
Q.19 The nature of cathode rays remains the same irrespective of the material used for
(a) gas (b) cathode
(c) glass (d) electrode
Q.20 Mass of electron is
(a) 9.1 x 10–31 kg (b) 9.109 x 10–32 gm
(c) 8.1 x 10–31 g (d) 9.1 x 10–31 mg
Q.21 The charge on an electron is
(a) 1.602 x 10–19 c (b) 1.602 x 10–18 c
(c) 1.602 x 10–19 c (d) 1.602 x 10–21 c
Q.22 The charge on the proton is
(a) + 1.602 x 10–19 c (b) zero
(c) – 1.602 x 1019 c (d) 1.602 x 10–19 c
Q.23 The charge on the neutron is
(a) 1.602 x 10–19 c (b) zero
(c) – 1.602 x 10–19 c (d) + 1.602 x 10–19 c
Q.24 The calculated e/m value of electron is
(a) 1.602 x 1019 c kg–1 (b) 1.7588 x 10–11 c kg–1
(c) 1.7588 x 10–13 c kg–1 (d) 1.759 x 109 c kg
Q.25 The mass of proton is
(a) 9.11 x 10–31 kg (b) 1.676 x 10–27 kg
(c) 1.60 x 10–19 kg (d) 1.675 x 10–27 kg
Q.26 The mass of neutron is
(a) 1.675 x 10–27 kg (b) 1.675 x 10–25 kg
(c) 9.11 x 10–31 kg (d) 1.60 x 10–19 kg
Q.27 The charge on electron was determined by
(a) J.J. Thomson (b) Millikan
(c) Rutherford (d) Bohr
Q.28 Alpha particles are identical to
(a) hydrogen atoms (b) helium atoms
(c) helium nuclei (d) fast moving electrons
Q.29 Bombardment of Beryllium with alpha particles generates
(a) proton (b) neutron
(c) electron (d) positron
Q.30 The colour of the glow produced in the discharge tube depends upon
(a) gas (b) electrodes
(c) composition of gas (d) pressure
Q.31 When the pressure of the gas in discharge tube is reduced, which of the following becomes more prominent
(a) gas glows (b) gas ionizes
(c) a discharge takes place (d) gas conducts electricity
Q.32 Goldstein discovered that besides the cathode rays, another type of rays are produced in the discharge tube which are called
(a) alpha rays (b) beta rays
(c) positive rays (d) gamma rays
Q.33 The e/m value for the positive rays in the discharge tube depends upon
(a) nature of electrode use
(b) nature of gas used
(c) composition of the gas
(d) pressure
Q.34 The distance between the two adjacent crests or troughs is called
(a) wave number (b) frequency
(c) wavelength (d) amplitude
Q.35 The value of Planck’s constant “h” is
(a) 6.625 x 10–34 cal (b) 6.625 x 10–34 J sec
(c) 6.625 x 10–34 kJ (d) 6.625 x 10–34 k cal
Q.36 In the Bohr’s model of atom the electron in an energy level emits or absorbs energy only when it
(a) remains in the same energy level
(b) dies out
(c) changes its energy level
(d) jumps away
Q.37 The energy associated with an electron resolving in first orbit is
(a) – 2.178 x 10–18 k J/mol
(b) – 1313.31 k J/mol
(c) – 328.32 k J/mol
(d) – 82.08 k J/mol
Q.38 The regions of spectrum are
(a) three (b) seven
(c) eight (d) five
Q.39 The dispersion of the components of white light when it is passed through prism is called
(a) rainbow (b) light pattern
(c) refraction (d) spectrum
Q.40 Which of the following colours has the shortest wavelength in the visible spectrum of light
(a) red (b) blue
(c) violet (d) green
Q.41 Which of the following colours has the longest wavelength in the visible spectrum of light
(a) red (b) blue
(c) violet (d) green
Q.42 A spectrum containing wavelength of all wavelengths is called
(a) continuous (b) discontinuous
(c) line (d) atomic
Q.43 A spectrum showing only certain colours of light is called
(a) continuous (b) line
(c) discontinuous (d) band
Q.44 The wavelength range of visible spectrum is
(a) 400–750 nm (b) 300–400 nm
(c) 350–600 nm (d) 200–400 nm
Q.45 The spectral lines of Lyman series (uv region) are produced when electron jumps from higher orbit to
(a) 1st orbit (b) 2nd orbit
(c) 3rd orbit (d) 4th orbit
Q.46 The spectral lines of Balmer series (visible region) are produced when electron jumps from higher orbit to
(a) 1st orbit (b) 2nd orbit
(c) 3rd orbit (d) 4th orbit
Q.47 The spectral lines of Paschen series (visible region) are produced when electron jumps from higher orbit to
(a) 1st orbit (b) 2nd orbit
(c) 3rd orbit (d) 4th orbit
Q.48 The spectral lines of Bracket series (visible region) are produced when electron jumps from higher orbit to
(a) 1st orbit (b) 2nd orbit
(c) 3rd orbit (d) 4th orbit
Q.49 A dual character of matter particles in motion was postulated by
(a) De–Broglie (b) Planck
(c) Einstein (d) Schrodinger
Q.50 If an electron is moving with a velocity of 2.188 x 106 m/s then its wavelength will be
(a) 0.33 x 106 nm (b) 0.33 x 10–2 nm
(c) 0.33 nm (d) 0.22 nm
Q.51 If a stone of 1gm is many with a velocity of 10m/s then its wavelength will be
(a) 6.65 x 10–30 m (b) 6.65 x 10–25 m
(c) 6.65 x 10–28 m (d) 6.65 x 10–12 m
Q.52 The space around the nucleus where the probability of finding the electron is maximum is called
(a) an orbital (b) an orbit
(c) energy level (d) a shell
Q.53 Which orbital has dumb–bell shape
(a) s–orbital (b) p–orbital
(c) d–orbital (d) f–orbital
Q.54 Which of the following quantum numbers describes energy of an electron in an atom
(a) principal quantum (b) azimuthal quantum
(c) magnetic quantum (d) spin quantum
Q.55 Which of the following quantum numbers describes shape of an electron in an atom
(a) principal quantum (b) azimuthal quantum
(c) magnetic quantum (d) spin quantum
Q.56 The degenerate orbital in p–subshell is
(a) 2 (b) 3
(c) 5 (d) 7
Q.57 When 4p orbital is complete the entering electron goes into
(a) 4d (b) 4f
(c) 5s (d) 5p
Q.58 x + l value for 3d will be
(a) 3 (b) 4
(c) 5 (d) 6
Q.59 Maximum number of electrons in 3f orbitals is
(a) 2 (b) zero
(c) 6 (d) 14
Q.60 Maximum number of electrons in M–shell is
(a) 2 (b) 8
(c) 18 (d) 32
Q.61 An orbital can have maximum electrons
(a) 2 (b) 8
(c) 18 (d) 6
Q.62 n + l value for 4f will
(a) 2 (b) 5
(c) 7 (d) 9
Q.63 When a spectrum of light is formed by the radiation given off by a substance it is called
(a) line spectrum (b) continuous spectrum
(c) emission spectrum (d) absorption spectrum
Q.64 Neutron was discovered by
(a) Chadwick (b) Bohr
(c) J.J. Thomson (d) Einstein
Q.65 Cathode rays can drive a small paddle wheel which shows that they
(a) are positively charged
(b) possess momentum
(c) do not possess momentum
(d) none of these
Q.66 Slow neutrons are generally more effective than fastness for the purpose of
(a) effusion (b) fission
(c) penetration (d) absorption
Q.67 The wavelength associated with the moving stone
(a) can be measured by many methods
(b) cannot be measured by any method
(c) can be measure by some method
(d) none of these
Q.68 Radius of orbit of an electron and velocity of electron are
(a) directly proportional to each other
(b) inversely proportional to each other
(c) independent to each other
(d) none of these
Q.69 The values of magnetic quantum number give us information about the number of orbitals in a
(a) small shell (b) orbit
(c) subshell (d) none of these
Q.70 Which of the following terms are used for the number of positive charges on the nucleus of an atom
(a) atomic number (b) atomic mass
(c) nuclear charge (d) atomic charge
Q.71 The uncertainty principle was stated by
(a) de Broglie (b) Heinsenberg
(c) Einstein (d) Schrodinger
Q.72 When a pressure in a discharge tube is reduced, which of the following phenomenon becomes very prominent
(a) gas conducts electricity
(b) a discharge takes place
(c) gas ionizes
(d) gas glows
Q.73 Atom bomb is based on the principle of
(a) nuclear fusion
(b) nuclear fission
(c) fusion and fission both
(d) radioactivity
Q.74 A spinning electron creates
(a) magnetic field (b) electric field
(c) quantum field (d) none of these
Q.75 The volume of space in which there is 95% chance of finding an electron is
(a) orbit (b) atomic orbital
(c) degenerate orbital (d) quantized orbital
Q.76 Planck’s equation is
(a) E = mc2 (b) E = hv
(c) E = hv2 (d) E = mc
Q.77 In an atom, the electrons
(a) are stationary in various energy levels
(b) are distributed in three dimensional charge cloud around the nucleus
(c) embedded in space around the nucleus
(d) revolve around the nucleus at random
Q.78 The mass number of an element is equal to
(a) number of electrons in an atom
(b) number of protons and neutrons in the nucleus
(c) number of protons in the nucleus
(d) number of neutrons in the nucleus
Q.79 The energy of bounded electron in H atom is
(a) positive (b) negative
(c) zero (d) none of these
Q.80 Quantum number which has symbol “n” is called
(a) principal quantum (b) Azimuthal quantum
(c) Spin quantum (d) Magnetic quantum
| Questions | 1 | 2 | 3 | 4 | 5 |
| Answers | b | a | c | a | c |
| Questions | 6 | 7 | 8 | 9 | 10 |
| Answers | c | a | c | c | c |
| Questions | 11 | 12 | 13 | 14 | 15 |
| Answers | a | c | d | b | c |
| Questions | 16 | 17 | 18 | 19 | 20 |
| Answers | b | c | b | d | a |
| Questions | 21 | 22 | 23 | 24 | 25 |
| Answers | c | a | b | b | b |
| Questions | 26 | 27 | 28 | 29 | 30 |
| Answers | a | b | c | b | c |
| Questions | 31 | 32 | 33 | 34 | 35 |
| Answers | c | c | b | c | b |
| Questions | 36 | 37 | 38 | 39 | 40 |
| Answers | c | b | c | d | c |
| Questions | 41 | 42 | 43 | 44 | 45 |
| Answers | a | a | b | a | a |
| Questions | 46 | 47 | 48 | 49 | 50 |
| Answers | b | c | d | a | c |
| Questions | 51 | 52 | 53 | 54 | 55 |
| Answers | a | a | b | a | b |
| Questions | 56 | 57 | 58 | 59 | 60 |
| Answers | b | c | c | b | c |
| Questions | 61 | 62 | 63 | 64 | 65 |
| Answers | a | c | c | a | b |
| Questions | 66 | 67 | 68 | 69 | 70 |
| Answers | b | b | b | c | a |
| Questions | 71 | 72 | 73 | 74 | 75 |
| Answers | b | b | b | a | b |
| Questions | 76 | 77 | 78 | 79 | 80 |
| Answers | b | b | b | b | a |
Q.1 Why it is necessary to decrease the pressure in the discharge tube to get the cathode rays?
Ans.
The current does not flow through the gas at ordinary pressure even at high voltage about 500 volts. However when the pressure inside the tube is decreased, the gas in the tube begins to conduct electricity at low pressure. Therefore it is necessary to decrease the pressure in the discharge tube to get the cathode rays.
Q.2 Which ever gas is used in the discharge tube the nature of the cathode rays remains the same why?
Ans.
A cathode ray consists of beam of electrons and electrons are constituents of all matter so, cathode rays do not depend upon the nature of the gas. Therefore, whichever gas is used in the discharge tube, the nature of cathode rays remains the same.
Q.3 Why e/m value of cathode rays is just equal to that of electrons?
Ans.
A cathode ray consists of beam of electrons, so cathode rays are actually electrons. Therefore e/m value of cathode ray is just equal to that of electron.
Q.4 The bending of the cathode rays in the electric and magnetic field show that they are negatively charged.
Ans.
The cathode ray beam travels in a straight line from the cathode to anode. The beam bends toward the south pole of the magnet when it passes through the magnetic field, which shows the cathode rays are negatively charged.
Q.5 Why positive rays are also called canal rays?
Ans.
Since positive rays produced in the discharge tube passed through the canals or holes of cathode, therefore positive rays are also called canal rays.
Q.6 The e/m values of positive rays for different gases are different but those for cathode rays, the e/m value is the same.
Ans.
The e/m value of positive rays depends upon the nature of gas used in the discharge tube. The characteristic of the gas varies from gas to gas, but for cathode rays e/m value is independent of the nature of the gas. Therefore, e/m values of positive rays for different gases are different but those for cathode rays the e/m value is the same.
Q.7 The e/m value for positive rays obtained from hydrogen gas 1836 times more than that of an electron?
Ans.
The mass of hydrogen gas is 1836 times more than that of an electron. Cathode rays consist of beam of electrons. The e/m value for positive rays depends upon the gas used in the tube, and e/m value for cathode rays is independent of the nature of the gas. Therefore e/m value for positive rays obtained from H2 gas is 1836 times less than that of cathode rays. Heavier the gas, the smaller the e/m value for positive rays.
Q.8 Justify, that cathode rays are material particles.
Ans.
Cathode rays drive a small paddle, wheel which shows that these rays posses momentum. From this observation, it is inferred that cathode rays are not rays but particles having a definite mass and velocity. Therefore cathode rays are material particles.
Q.9 How neutrons are produced?
Ans.
When a stream of a–particles from a polonium source is directed at beryllium target, penetrating radiations are produced, which are called neutrons.
He + Be ® C + n
Q.10 Why the neutrons are used as projectile?
Ans:
The particles, which hit the nucleus and can change its nature are called projectile. A projectile must be chargeless otherwise it will be captured or repelled by the nucleus. The slow moving neutrons cause nuclear reactions like fission and are used in artificial radioactivity. They are chargeless; therefore they can be used as projectile in nuclear research.
n + 
Cu → 
Cu + hv (γ- radiations)
Cu → Cu + hv (γ- radiations)
Cu → 
Zn +
-1e (β-particle)Q.10 How are x–rays produced?
Ans.
X–rays are produced when fast moving electrons collide with heavy metal anode in the discharge tube.
Q.11 why the potential energy of bounded electron is negative in Bohr’s model?
Ans.
The potential energy of bounded electron is negative, because the energy of separated nucleus and electron is taken to be zero. As electron is brought from infinity towards the nucleus to form a stable state of the atom, energy is released because of attractive forces and the energy becomes less than zero, or negative. Therefore, the energy of the bounded electron is negative.
Q.12 Why the total energy of bounded electron in negative in Bohr’s model?
Ans.
The total energy of bounded electron is negative because the electron is under the force of attraction of the nucleus to have a stable state of the atom. More over when we calculate the total energy of the bounded electron, which is the sum of K.E. and P. E comes which is also negative.
Q.13 Explain that energy of an electron is inversely proportional to n2, but energy of higher orbits are always greater than those of the lower orbits in Bohr’s model.
Ans.
The energy of an electron in the nth orbit is
En = –
where e, m, 00 and h are all constants, thus En µ
The more negative the energy is the more stable will be the atom. The energy becomes successively less negative, therefore the energy values of higher orbits are always greater than those of the lower orbits.
Q.14 Explain the energy difference between adjacent levels goes on decreasing sharply in Bohr’s model.
Ans.
The energy difference between adjacent levels goes on decreasing, because the distance between the adjacent orbits increases.
Q.15 why does cathode rays produce shadow of an opaque object placed in their path.
Ans.
Any object which is material in nature, produces its shadow. Since cathode rays are material in nature, therefore, they produce shadow of an opaque object placed in their path.
Q.16 Give the main points of quantum theory of radiation.
Ans.
1. Energy is emitted or absorbed by atoms only in the form of packets called quantum.
2. The amount of energy associated with a quantum of radiation is proportional to the frequency (u) of the radiation.
E µ u
or E = hu
3. A body can emit or absorb energy only in terms of integral multiples of quantum.
E = nhu (where n = 1, 2, 3, 4, 5, ……..)
Q.17 Define frequency, wavelength and wave number.
Ans. Frequency (u):
The number of waves passing through a point per second is called frequency (u). Its units are cycles s–1.
Wavelength (l):
The distance between two successive crests or troughs is called wavelength “l” and is expressed in Ao or nm.
Wave number:
The number of waves per unit length is called wave number and is reciprocal of wave length.
=
The wave number is expressed (m–1) or per meter.
Q.18 What is spectrum? Differentiate between continuous spectrum and line spectrum.
Ans.
The dispersion of the components of white light, when it is passed through prism is called spectrum. The distribution among various wavelengths of the radiant energy emitted or absorbed by an object is also called spectrum.
Continuous spectrum:
A spectrum containing light of all wavelengths is called continuous spectrum.
In this type of spectrum, the boundary line between the colours cannot be marked. The colours diffuse into each other. One colour merges into another without any dark space. The best example of continuous spectrum is rainbow.
Line spectrum:
When an element or its compound is volatilized on a flame and the light emitted is seen through, a spectrometer. We see distinct lines separated by dark spaces. This type of spectrum is called line spectrum. This is the characteristic of an atom.
Q.19 Describe briefly Rutherford ’s atomic model.
Ans.
According to Rutherford ’s model most of the mass of the atom (99.95%) is concentrated in a positively charged centre, called nucleus around which the negatively charged electrons move.
Q.20 On which experiment Rutherford ’s atomic model is based on, describe it briefly?
Ans.
Q.21 Define orbit and orbital.
Ans. Orbit:
A definite circular path at a definite distance from the nucleus in which the electrons revolve around the nucleus is called an orbit.
K, L, M, N are orbits.
Orbital:
A three dimensional region or space around the nucleus, within which the probability of finding an electron is maximum called an orbital, s, p, d and f are atomic orbitals.
Q.22 What do you understand by wave particle duality and what is the de Broglei relation?
Ans.
According to de Broglei, all matter particles in motion have a dual character. It means that electrons, protons, neutrons, atoms, and molecules, possess the characteristics of both the material particle and a wave. This is called wave particle duality in matter.
De Broglei derived a mathematical equation which relates the wavelength (l) of the electron to the momentum of electron (mv)
l =
Where l = wavelength v = velocity of electron
M = mass of electron and h is Planck’s constant.
This equation l = is called de Broglie relation.
Q.23 What is Heisenberg’s uncertainty principles?
Ans.
Heisenberg showed that it is impossible to determine simultaneously both the position and momentum of an electron. Suppose that Dx is the uncertainty in the measurement of the position and Dp is the uncertainty in the measurement of momentum of an electron.
Dx . Dp ³
This relationship is called uncertainty principle.
Q.24 What are quantum numbers?
Ans.
The dimensionless numbers, rise naturally when the Schrodinger wave equation is solved for electron wave patterns and their energies are called quantum numbers.
These numbers describe the behaviour of electron in an atom completely.
There are four quantum numbers.
1. Principal quantum number “n”
It describes the energy of an electron in an atom. The value of n represents the shell or energy level in which the electron revolves around the nucleus. These shells are named as K, L, M, N, O, P, having the values of n, 1, 2, 3, 4, 5 and 6 respectively. The greater the value of n, the greater will be the distance from the nucleus and greater will be the energy of electron in the shell.
2. Azimuthal quantum number “l”
It determines the shape of orbital, it can have any integer value from 0 to n–l. this quantum number is used to represent the sub–shells, and these value are l = 0, 1, 2, 3. These values represent different sub–shells which are designated as s, p, d, and f, with values of l = 0, 1, 2, 3 respectively.
3. Magnetic quantum number (m)
It describes the orientation of the orbital in space. It can have all the integral values between + l and – l through zero i.e. + l …….. 0 …….. – l. For each value of l, there will be
(2l + 1) values of m. actually the values of m gives us the information of degeneracy of orbitals in space.
(2l + 1) values of m. actually the values of m gives us the information of degeneracy of orbitals in space.
4. Spin quantum number (s)
It describes the spin of electron in atom. Since an electron can spin clockwise or anti clockwise, thus two possible values are + and – depending upon the spin of electron.
Q.25 What is n + l rule?
Ans.
This rule says that sub–shells are arranged in the increasing order of (n + l) values and if any two sub–shells have the same (n + l) values, then the sub–shell is filled first whose n values is smaller.
Q.26 What is the origin of line spectrum?
Ans.
According to Bohr’s theory each bright line in a line spectrum results from the downward jump of electron from a higher energy E2 to lower energy E1. This difference in energy (E2 – E1) is emitted as radiation of definite frequency in the form of spectral line.
According to the quantum theory of radiation,
E1 – E2 = hu
Or u =
Q.27 When is Zeeman effect?
Ans.
When the excited atoms of hydrogen are placed in a magnetic field, its spectral line are further split up in to closely spaced lines. This type of splitting of spectral lines is called Zeeman effect.
Q.28 What is stark effect?
Ans.
When the excited hydrogen atom are placed in an electric field, its spectral lines are further split up into closely spaced lines. This type of splitting of spectral lines is called stark effect.
Q.29 What is Mosely’s Law?
Ans.
Mosely’s law states that the frequency of spectral line in
x–ray spectrum varies as the square of atomic number of an element emitting it. This law convinces us that it is the atomic number and not the atomic mass of the element which determines its characteristic properties, both physical and chemical.
x–ray spectrum varies as the square of atomic number of an element emitting it. This law convinces us that it is the atomic number and not the atomic mass of the element which determines its characteristic properties, both physical and chemical.
Q.30 Describe Summerfield’s modification of Bohr’s model atom.
Ans.
Summerfield suggested that the moving electron revolves in elliptical orbits in addition to circular orbit, with the nucleus situated at one of the foci of the ellipse. The elliptical paths of the moving electron go on changing their position in space, and the nucleus is buried by the electronic cloud from all the sides.
Q.31 Which of these orbitals, 3d or 4s has higher energy level?
Ans.
For 3d, n + l = 3 + 2 = 5 and for 4s, n + l = 4 + 0 = 4. Therefore 3d orbital has higher energy, than 4s orbital.
Q.32 How many maximum number of electron can have an orbital and a shell?
Ans.
An orbital can have maximum two electrons with opposite spins. A shell can have maximum of 2n2 electrons, where “n” is the principal quantum number. First shell can have maximum 2 electrons, 2nd shell have 8 electrons 3rd shell have 18 electrons etc.
Q.33 Distribute electrons in orbitals of 19K, 29Cu, 24Cr, 53I.
Ans.
19K ® 1s2, 2s2, 2p6, 3s2, 3p6, 4s1
29Cu ® 1s2, 2s2, 2p6, 3s2 3p6, 3d10, 4s1
24Cr ® 1s2, 2s2, 2p6, 3s2 3p6, 3d5, 4s1
53I ® 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2, 5p5
Q.34 What does it mean, when we say energy is quantized?
Ans.
Quantization means that energy can only be absorbed or emitted in specific amounts or multiples of these amounts. This minimum amount of energy is equal to a constant times the frequency of the radiation absorbed or emitted E = hv.
Q.35 Why do not we notice the quantization of energy in every day activities?
Ans.
In everyday activities, macroscopic objects such as our bodies gain or lose total amounts of energy much larger than a single quantum, hv. The gain or loss of the relatively minuscule quantum of energy in unnoticed.
Q.36 Explain the existence of line spectra is consistent with Bohr’s theory of quantized energies for the electron in the hydrogen atom.
Ans.
When applied to atoms, the notion of quantized energies means that only certain values of D E are allowed. These are represented by the lines in the emission spectra of excited atoms.
Q.37 In what ways does de Broglie’s hypothesis require revision of our picture of the H–atom based on Bohr’s model?
Ans.
De Broglie’s hypothesis not electrons have a characteristic wavelength requires, revision of Bohr’s particle only model. For example the idea of a fixed orbit for the electron in hydrogen is hard, to reconcile with the wave properties of electron.
Q.38 (a) For n = 4 what are possible values of l?
(b) For l = 2 what are the possible values of m.
Ans.
(a) n = 4 l = 3, 2, 1, 0
(b) l = 2 m = – 2, – 1, 0, 1, 2
Q.39 Which of the following are permissible sets of quantum numbers for an electron in a hydrogen atom?
(a) n = 2 l = 1 m = 1
(b) n = 1 l = 0 m = – 1
(c) n = 4 l = 2 m = – 2
(d) n = 3 l = 3 m = 0
Ans.
(a) permissible 2p (b) not permissible
(c) Permissible 4d (d) not permissible
Q.40 (a) What are the possible values of the electron spin quantum numbers?
(b) What piece of experimental equipment can be used to distinguish electrons that have different values of the electron spin quantum number?
(c) Two electrons in an atom both occupy the Is orbital. What quantity must be different for the two electrons? What principle governs the answer to this question?
Ans.
(a) + , –
(b) A magnet with a strong inhomogeneous magnetic field.
(c) They must have different spin quantum number values. The Pauli exclusion principle.
Q.41 Give region of different spectral lines.
Ans.
1. Lyman series (U. V. region)
2. Balmer series (visible region)
3. Paschen series (I. R. region)
4. Bracket series (I. R. region)
5. Pfund series (I. R. region)
TEXT BOOK EXERCISE
Q1. Select the most suitable answer for the given one.
i. The nature of the positive rays depends on
(a) The nature of the electrode
(b) The nature of the discharge tube
(c) The nature of the residual gas
(d) all of the above
ii. The velocity of photon is
(a) Independent of its wavelength
(b) Depends on its wavelength
(c) Equal to square of its amplitude
(d) Depends on its source
iii. The wave number of the light emitted by a certain source is 2 x 106 m. the wavelength of this light will be
(a) 500nm (b) 500m
(c) 200nm (d) 5 x 107 m
iv. Rutherford ’s model of an atom failed because
(a) The atom did not have a nucleus an electron
(b) It did not account for the attraction between protons and neutrons
(c) It did not account for the stablility of the atom
(d) There is actually no space between the nucleus and the electrons
v. Bohr model of atom is contradicted by
(a) Planck quantum theory
(b) Pauli’s exclusion principle
(c) Heisenberg’s uncertainty principle
(d) all of the above
vi. Splitting of spectral lines when atoms are subjected to strong electric field is called.
(a) Zeeman effect (b) Stark effect
(c) Photoelectric effect (d) Compton Effect
vii. In the ground state of an atom, the electron is present
(a) In the nucleus (b) in the second shell
(c) Nearest to the nucleus
(d) farthest from the nucleus
viii. Quantum number values for 2p orbitals are
(a) n=2, l=l (b) n=1, l=2
(c) n=1, l=0 (d) n=2, l=0
ix. Orbitals having same energy are called
(a) Hybrid orbitals (b) valence orbitals
(c) Degenerate orbitals (d) d-orbitals
x. when 6d orbitals is complete, the entering electron goes into
(a) 7f (b) 7s (c) 7p (d) 7d
Ans. (i)c (ii)a (iii)a (iv)c (v)c
(vi)b (vii)c (viii)a (ix)c (x)c
Q.2 Fill in the blanks with suitable words
(i) B-particles are nothing but _______moving with a very high speed.
(ii) The charge on one mole of electrons is ________coulombs.
(iii) The mass of hydrogen atom is _________grams.
(iv) The mass of one mole of electron is _________.
(v) Energy is ________when electron jumps from higher to a lower orbit.
(vi) The ionization energy of hydrogen atom can be given by formula ________.
(vii) For d sub-shell, the azimuthal quantum number has a value of ________.
(viii) The number of electrons in a given sub-shell is given by formula ________.
(ix) The electronic configuration of H- is ________.
Ans. i)electrons ii)96500 iii)1066x10-24 iv)5.484x10-7kg v)emitted vi) Bohr vii)2 viii)2(2l+1) ix) 1s2
Q.3 Indicate true or false as the case may be.
(i) A neutron is slightly lighter particle than a proton.
(ii) A photon is the massless bundle of energy but has momentum.
(iii) The unit of Rydberg constant is the reciprocal of unit of length.
(iv) The actual isotopic mass is a whole number.
(v) Heisenberg’s uncertainty principle is applicable to macroscopic bodies.
(vi) The nodal plane in an orbital is the plane of zero electron density.
(vii) The number of orbitals present in a sublevel is given by the formula (2l_1)
(viii) Magnetic quantum number was introduced to explain Zeeman and stark effects.
(ix) Spin quantum number tells us the direction of spin of electron around the nucleus.
Ans. i)False ii)True iii)True iv)False v)False vi)True vii)True viii)True ix)Ture
Q.4 Keeping in mind the discharge tube experiment , answer the following questions.
(a) Why is it necessary to decrease the pressure in the discharge tube to get the cathode rays?
Ans. There will be no flow of current through the gas when the pressure in the discharge tube is high. In the presence of high pressure, the cathode rays will not flow from the cathode surface.
(b) Which ever gas is used in the discharge tube the nature of the cathode rays remains the same. Why?
Ans. Cathode rays are composed of negatively charged particles (electrons). They are constituents of all gases. So, cathode rays are independent of the nature of the gas in the discharge tube.
(c) Why e/m value of the cathode rays is equal to that of electron?
Ans. Cathode rays are composed of electrons, so their e/m value is just equal to that of electron.
(d) How the bending of the cathode rays in the electric and magnetic fields shows that they are negatively charged?
Ans. when cathode rays are passed through an electric field created by two charged metal plates, they are deflected towards the positively charged plate. This shows that they are negatively charged.
When cathode rays are passed between the poles of a magnet, the magnet neither attracts nor repels but cause them to move in a curved path perpendicular to the line drawn between the poles of the magnet. This shows that they are negatively charged.
(e) Why positive rays are also called canal rays?
Ans. Positive rays pass through canals or holes in the cathode, so they are called canal rays.
(f) The e/m value of positive rays for different gases are different but those for cathode rays the e/m values is the same Justify it.
Ans. The e/m value for positive rays are different for different gases because they differ in mass. the mass of the positive particles is the same as that of the atom or molecule form which it is created. Heavier the gas, smaller the e/m value.
The e/m value for cathode rays is the same because cathode rays are composed of electrons which have constant mass.
(g) The e/m value for positive rays obtained from hydrogen gas is 1836 time less than that of cathode rays. Justify it.
Ans. The e/m value for positive rays obtained form hydrogen gas is 1836 time less than that of cathode rays. This is because the mass of proton which is created from H-atom is 1836 time more than that of an electron (cathode rays particle).
Q.5 (a) Explain Milliken’s oil drop experiment to determine the charge of an electron.
(b) What is J.J Thomson’s experiment for determining e/m value of electron?
(c) Evaluate mass of electron from the above two experiments.
Q.6 (a) Discuss Chadwick’s experiment for the discovery of neutrons. Compare the properties of electron, proton, and neutron.
(b) Rutherford ’s atomic model is based on the scattering of a-particles from a thin gold foil. Discuss it and explain the conclusions.
Q.7 (a) Give the postulates of Bohr’s atomic model. Which postulate tell us that orbits are stationary and energy is quantized?
(b) Derive the equation for the radius of nth orbit of hydrogen atom using Bohr’s model.
(c) How does the above equation tell you that?
(i) Radius is directly proportional to the square of the number of orbit.
(ii) Radius is inversely proportional to the number of proton in the nucleus.
Ans. The equation for the radius of nth orbit is:
r=
Since, , h, ,m and e are constant , therefore , the factoris constant.
, h, ,m and e are constant , therefore , the factoris constant.Therefore, r=constant x
Or
r =
Hence, we can say:
(i) Radius is directly proportional to the square of the number of orbit.
(ii) Radius is inversely proportional to the number of protons in the nucleus
(d) How do you come to know that the velocities of electrons in higher orbits are less than those in lower orbits of hydrogen atom?
Ans. According to Bohr, since the electron keeps on revolving around the nucleus:
Therefore, centrifugal force=Electrostatic force of attraction
=
or r=
For H-atom,Z=1
r= Since, e,,and m are constant, therefore, the factor 
is constant.
,and m are constant, therefore, the factor is constant.Therefore, r=constant x 
r=
Hence, the radius of a moving electron is inversely proportional to the square of its velocity. The smaller the radius of the orbit, the higher is the velocity of electron. Hence, the velocities of electrons in higher orbits are less than those in lower orbits of hydrogen atom.
(e) Justify that the distance gaps between different orbits go on increasing form the lower.
Ans. We know that: r=0.529 
x[n2]
x[n2] When n=1 r1=0.529
When n=2 r2=0.529x4=2.11
x4=2.11 When n=3 r3=0.529x9=4.75
x9=4.75 When n=4 r4=0.529x16=8.4
x16=8.4 When n=5 r5=0.529x25=13.22
x25=13.22 Distance between orbits are:
r2 –r1=(2.11 – 0.529) =1.581
=1.581 r3 –r2=(4.75 – 2.11) =2.64
=2.64 r4 –r3=(8.4 – 4.75) =3.65
=3.65 r5 –r4=(13.22 – 8.4) =4.82
=4.82 From the data of radius difference, it is clear that the distance gaps between different orbits go on increasing from the lower to the higher orbits.
Q8. Derive the formula for calculating the energy of an electron in nth orbit using Bohr’s model. Keeping in view this formula explain the following:
(a) The potential energy of the bounded electron is negative.
Ans. According to Bohr, the energy of electron is calculated from the equation:
En=- 2.178 x 10-18[]
When n=, then En=0
, then En=0 Consider that an electron is present at an infinite distance from the nucleus, so there is no interaction between the two. The energy of this electron is zero.
Now, suppose that the electron moves closer and closer to the nucleus. Since electron is negatively charged and nucleus is positively charged, no work needs to be done on the electron. The electron can move towards the nucleus by itself due to electrical attraction. Thus, work is done by the electron itself as it moves towards the nucleus. As a result, the potential energy falls, i.e. it become less than zero. Any value less than zero is negative. Hence, the potential energy of electron becomes more and more negative as the electron moves closer and closer towards the nucleus.
(b) Total energy of the bounded electron is negative.
Ans. When the electron is at infinite distance from the nucleus, there is no electrostatic interaction between the two. Therefore, the energy of the system in this state is assumed to be zero. As the electron moves closer to the nucleus, it does some work and energy is of the electron becomes negative. The negative value of energy would keep increasing as the electron moves to the energy levels nearer to the nucleus. the negative value of total energy shows that electron is bound by the nucleus i.e, electron is under the force of attraction of the nucleus.
(c) Energy of an electron is inversely proportional to n2 , but energy of higher orbits are always greater than those of the lower orbits.
Ans. Energy of electron:
The energy of electron in different orbits can be calculated by using the following equation:
En = - kJmol-1
Energy of an electron is inversely proportional to n2.
When n =1 E1=
=1312.36kJ mol-1
=1312.36kJ mol-1 When n =2 E2=
=328.09kJ mol-1
=328.09kJ mol-1 When n =3 E3=
=145.82kJ mol-1
=145.82kJ mol-1 When n =4 E4=
=82.023kJ mol-1
=82.023kJ mol-1 When n =5 E1=
=52.49kJ mol-1
=52.49kJ mol-1 The energy of an electron is inversely proportional to n2 .
As the value of ‘n’ increases, the value of energy increases. The energy of higher orbits are always greater than those of the lower orbits.
E5 >E4 >E3> E2 >E1
(d) The energy difference between adjacent levels goes on decreasing sharply.
Ans. The energy difference between adjacent levels can be found as:
E=E2 – E1 =(- 328.09) – (-1312.36) =984.27kJmol-1 E=E3 – E2 =(- 145.82) – (- 328.09) =182.27kJmol-1 E=E4 – E3 =(- 82.023) – (- 145.82) =63.797kJmol-1E=E5 – E4 =(- 52.49) – (- 82.023) =29.533kJmol-1 From the data of energy difference, it is found that the energy difference between adjacent levels goes on decreasing sharply.
Q9. (a) Derive the following equation for hydrogen atom which are related to:
i. Energy difference between two levels, n1 and n2 .
ii. Frequency of photon emitted when an electron jumps from n2 to n1 .
iii. Wave number of the photon when the electron jumps from n2 to n1 .
(b) Justify that Bohr’s equation for the wave number can explain the spectral lines of Lyman, Blamer and paschen series.
Q10. (a) What is spectrum? Differentiate between continuous spectrum, and line spectrum.
(b) Comparison between line emission and line absorption spectra.
Ans.
| Line emission spectrum | Line absorption spectrum |
| 1. “An atomic spectrum which consists of bright lines against a dark background is called line emission spectrum. “ 2. it is produced when radiations emitted by an excited substance are analysed in a spectroscope. | 1. “An atomic spectrum which consists of bright lines against a dark background is called line emission spectrum. “ 2. it is produced when white is passed through the gaseous element and the transmitted rays are analysed in a spectroscope. |
(c) What is the origin of the line spectrum?
Q11. (a) Hydrogen atom and He+ are monoelectronic system, but the size of He+ is much smaller than H+ , why?
Ans. H-atom and He+ are monoelectronic system. It means both H-atom and He+ have one electron in the valence shell. H-atom has one proton in the nucleus whereas He+ has two proton in the nucleus. So, the force of attraction between two protons and one electron is greater than one proton and one electron. Hence, the size of He+ is much smaller than H-atom.
Also, we know that: r=0.529
The size of H-atom: r=0.529
=0.2645
=0.2645 The size of He+ ion: r=0.529
=0.529 x =0.2645
=0.529 x =0.2645 Hence, the size of He+ is much smaller than H-atom. This is because the nucleus of He+ has greater attraction for the electron as compared to H-atom which contains one proton in the nucleus.
(b) Do you think that the size of li-2+ is even smaller than HE+ ? Justify with calculation.
Ans. The size of He+ ion: r=0.529=0.2645
=0.2645 The size of li-2+ ion: r=0.529
=0.1763
=0.1763 The size of li-2+ ion is much smaller than the size of He+ ion:
Q12. (a) What are X-rays? What is their origin? How was the idea of atomic number derived from the discovery of X-rays?
(b) How does the Bohr’s model Justify the Moseley‘s equation?
Q13. Point out the defects of Bohr’s model. How these defects are partially covered by dual nature of electron and Heisenberg’s uncertainty principle?
Q14. (a) Briefly discuss the wave mechanical model of atom. How has it given the idea of orbital. Compare orbit and orbital.
Ans. Comparison between orbit and orbital:
| Orbit | Orbital |
| 1. “A circular path around the nucleus in which the electron revolves is called an orbit.” 2. It is circular in Shape. 3. It represents that an electron moves around the nucleus in one plane, i.e., in a flat surface. 4. It is against Heisenberg’s uncertainty principle. 5. The maximum number of electrons in an orbit is 2n2, where ‘n’ is the number of the orbit. | 1. “The volume of space within an atom in which there is 95% chance of finding an electron is called orbital.” 2. It may be spherical, dumbbell or double dumbbell in shape. 3. It represent that an electron can move around the nucleus in three dimensional space. 4. It is in accordance with Heisenberg’s uncertainty principle. 5. The maximum number of electrons in an orbital is two. |
(b) What are quantum number? Discuss their significance.
(c) When azimuthal quantum number has a value 3, then there are seven value of magnetic quantum number. Give reasons.
Q15. Discuss rules for the distribution of electrons in energy sub-shells and in orbitals. (a) What is (n+ l ) rule. Arrange the orbitals according to this rule. Do you think that this rule is applicable to degenerate orbitals?
(b) Distribute electrons in orbitals of 57La, 29Cu, 79Au, 24Cr, 53I, 86Rn.
Ans. Electronic configurations:
57La=1s2 2s2 2p6 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f4 (Actual configuration)
=1s2 2s2 2p6 3p6 4s2 3d10 4p6 5s2 4d10 5s2 5p6 5d1(Expected configuration)
29Cu =1s2 2s2 2p6 3d9 4s2 (Actual configuration)
=1s2 2s2 2p6 3p6 4s2 3d10 4s1 (Expected configuration)
79Au=1s2 2s2 2p6 3p6 4s2 3d10 4p6 5s2 4d10 4f145s2 5p6 5d10 gs1
24Cr=1s2 2s2 2p6 3s2 3p6 3d4 4s2 (Actual configuration)
=1s2 2s2 2p6 3s2 3p6 3d4 4s1 (Expected configuration)
53I=1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p5
86Rn==1s2 2s2 2p6 3s2 3p6 3d4 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6
From the above configuration, it is important to note that there are three irregularities in the general trend. The electronic configuration of Cr and Cu show deviation from the expected configuration.
Expected Configurations: Cr=1s2 2s2 2p6 3s2 3p6 3d4 4s1
Cu=1s2 2s2 2p6 3s2 3p6 3d4 4s1
Actual Configurations: Cr=1s2 2s2 2p6 3s2 3p6 3d4 4s2
Cu=1s2 2s2 2p6 3s2 3p6 3d4 4s2
This is because the half filled and fully filled configurations (i.e.,d5,d10,f7,f14 )have lower energy ot more stability. Thus, in order to become more stable, one of 4s electrons goes into 3d orbitals so that 3d orbitals get half filled or fully filled configuration in Cr and Cu respectively.
Reasons for stability of Half filled and fully filled orbitals:
1. Exchange energy; All the orbitals in a given sub-shell have equal energies. The electrons present in different orbitals of the same sub-shell can exchange their positions. Such an exchange of electrons results in release of energy called exchange energy . Greater the exchange energy, more is the stability associated with the orbitals. It has been observed that in case of exactly half-filled and fully filled orbitals the electrons can exchange their positions more readily as compared to other configurations. As a result, half-filled and fully filled configurations are more stable.
2. Symmetry: If removal or addition of an electron results in the symmetrical distribution of electrons in an orbital, the electronic configuration becomes more stable. Therefore, half-filled and fully filled configurations in which each orbital contains one and two electrons respectively , are more stable.
Irregular Structure of La: The electronic configuration of La is irregular and does no t follow the general trend. This is because strong nuclear attraction and less shielding effect caused by d and f electrons. In this case remember before adding any electron in the 4f orbital, a single electron is added to a 5d orbital. The remaining nine electrons enter the 5d sub-shell after the 4f sub-shell has been completely filled with fourteen electrons. Similarly, one electron enters the 6d sub-shell before any electron enters the 5f sub-shell.
Q16. Draw the shapes of s, p and d-orbitals. Justify these by keeping in view the azimuthal and magnetic quantum numbers.
Q17. A photon of light with energy 10-19 j is emitted by a source of light/
(a) Convert this energy into the wavelength, frequency and wave number of the photon in terms of meters, hertz and m-1 respectively.
Solution:
E=10-19 J h=6.625 x 10-34 js
Formula used: E=hv
V=
V=
V=1.51 x 1014 s-1
V=1.51 x 1014 Hz [CPS=s-1 =Hz]
Now, 
=
= E=10-19 J=10-19 kg m2 s-2 [1 J=1 kf m2 s-2 ]
h =6.625 x 10-34 kg m2 s-2 s
=6.625 x 10-34 kg m2s-1
= 
=19.875x 10-7 m=1.9875 x 10-6 m=1.9875 x 10-4 cm Now, 
=0.51 x 106 m-1=5.1 x 105 m-1 =5.1 x 103 cm-1(b) Convert this energy of the photon into ergs and calculate the wave length in cm, frequency in Hz and wave number in cm-1 .
h=6.625x 10-34 js or 6.625 x 10-27 ergs. C=3 x 108 ms-1 or 3x 10 +10 cms-1 .
Solution:
E= 10-19 J =10-19 x 107 erg [1 j =107 erg]
=10-12 erg
=1.98 x 10-6 m=1.98 x10-6 x 102 cm=1.98 x 10-4 cm=5x105 m-1 =5 x 105 x 10-2 cm-1 =5x 103 cm-1 Q18. The formula for calculating the energy of an electron in hydrogen atom gives by Bohr, s model
En =
Calculate the energy of the electron in first orbit of hydrogen atom. The values of various parameters are same as provided in Q.19.
Solution:
n = E=?
Formula: En =-2.178 x 10-18 [
]J
]J En=-2.178 x 10-18 [
]J
]J E=-2.178 x 10-18 J
E=-2.18 x 10-18 J
Q19. Bohr’s equation for the radius of nth orbit of electron in hydrogen atom is
rn =
(a) When the electron moves from n=1 to n=2, how much does the radius of the orbit increases.
Solution:
=8.85 x10-12 C2 J-1 m-1 ; m=9.108 x 10-31 kg h=6.624 x 10-34 js ; e=1.602 x 10-19 C ;=3.14
=3.14Form n=1 to n=2 : J=kg m2 s-2 : C=
s-1
s-1 Formula: rn =
r1= 
r1=
r1=
r1= 5.29 x 10-11 jm-1 s2 kg-1
r1= 5.29 x 10-11 kg m2 s2m-1 s2 kg-1
r1= 5.29 x 10-11 m
r1= 5.29 x 10-1
r1= 0.529
Also r2= 0.529 [22]
[22] R2= 0.529 x 4 =2.116
x 4 =2.116Increase in radius, (r2 - r1) =2.116- 0.529
- 0.529 =1.587
(b) What is the distance traveled by th electron when it goes from n=2 to n =3and n =9 to n=10?
=8.85x 10-12 C2 J-1 m-1 , h=6.24x10-34 js, =3.14, m=9.108 x10-31 kg , C=1.602 x 10-19 C
While doing calculations take care of units of energy parameter.
J=kgm2 s-2 , c=kg
m
s-1
m s-1Solution: r=0.529 (n2)
(n2) For n=2 r2 =0.529(22) =2.116
(22) =2.116 n =3 r3=0.529(32) =4.761
(32) =4.761 Distance traveled by the electron when it goes from n=9 to n =3
r3 – r2 =4.761 - 2.116 =Answer
- 2.116 =AnswerFor n=9 r9 =0529(92)=42.849
n =10 r10=0.529 (102)=52.9
(102)=52.9Distance traveled by the electron when it goes grom n=9 to n=10
r10 – r9 =52.9 - 42.849 =10.051 Answer
- 42.849 =10.051 AnswerQ20. Answer the following questions, by performing the calculation s.
(a) Calculate the energy of first five orbits of hydrogen atom an determine the energy differences between them.
(b) Justify that energy difference between second and third orbits is approximately five times smaller than that between first and second orbits.
(c) Calculate the energy of electron in He+ in first five orbits and justify that the energy differences are different from those of hydrogen atom.
(d) Do you think that group of the spectral lines of He+ are at different places than those for hydrogen atom? Give reasons.
Q21. Calculate the value of principal quantum number if an electron in hydrogen atom revolves in an orbit of energy - 0.242 x 10-18 j.
Solution:
E= - 242 x 10-18 j ; n=?
Formula: E=- 2.178 x 10-18
n2= - 2.178 x 10-18 x
n2=- 2.178 x 10-18x
n2=
n2=9
n= 3 Answer
Q22. Bohr’s formula for the energy levels of hydrogen atom for any system say H, He+, Li-2+ , etc is
En=
Or
En= - k [
]
] For hydrogen Z=1 and for He+ , Z=2
(a) Draw an energy level diagram for hydrogen atom and He+ .
(b) Thinking that k = 2.18 x 10-18 j, calculate the energy needed to remove the electron from hydrogen atom and from He+ .
Solution:
For H; Z=1 ; n=1 ; k=2.18 x 10-18 j
Formula En =-k[]
] E1 = -2.18 x 10-18 [
]
] E1=-2.18 x 10-18 j
=-2.18 x 10-18 [
] J=-2.18 x 10-18 0j=0 Now energy required to remove an electron from the orbit:
= - E1=0 – (-8.72 x10-18 J)=8.72 x 10-18 J Answer(c) How do you justify that the energies calculated in (b) are the ionization energies of H and He+ ?
Ans. The energy difference between first and infinite levels of energy for H atom is 2.18x 10-21 kJ and for He+ ion is 8.72x10-21 kJ are the ionization energies of H and He+ respectively. These values are the same as determined experimentally.
(c) Use Avogadro’s number to convert ionization energy values in kJ mol-1 or H and He+ .
Solution:
For H: =2.18 x 10-18 J
=2.18 x 10-18 J The value of energy obtained for the electron is in J/atom. If this quantity is multiplied by Avogadro’s number and divided by 1000, the value of E will become.
= 
=13.1236 x 102 kJ mol-1=1312.36 kJ mol-1 Answer For He+ ; =8.72 x 10-18 J/atom=8.72 x 10-18 x
=52.4944 x 102 kJ mol-1=5249.44 kJ mol-1 Answer (e) The experimental values of ionization energy of H and He+ are 1331 kJ mol-1 and 5250 kJ mol-1 respectively. How do you compare your values with experimental values ?
Q23. Calculate the wave number of the photon when the electron jumps from
i. n=5 to n=2.
ii. n=5 to n=1.]
In which series of spectral lines and spectral regions these photons will appear.
Solution:
(i) n=5 to n =1 . =?
=?Formula : =1.09678 x 107 [
-
]m-1
=1.09678 x 107 [-]m-1=1.09678x 107[
-
]m-1=1.09678x 107[
]m-1=1.09678x 107[
]m-1=1.09678x 107m-1 =2.3x106 m-1 Answer The photon will appear in the Balmer series
ii) n =5 to n= 1
Solution:
n = 5 to n=1 ; =?
=? Formula: =1.09678 x 107 [-]m-1
=1.09678 x 107 [-]m-1=1.09678 x 107 [
-
]m-1=1.09678 x 107 [
-]m-1=1.09678 x 107 [
]m-1=1.09678 x 107 m-1 Answer The photon will appear in the Lyman series.
Q24. A photon of a wave number 102.70 x 105 m-1 is emitted when electron jumps form higher to n=1.
(a) Determine the number of that orbit from where the electron falls.
Solution:
=102.70 x 105 ; n2=?Formula used: =1.09678 x 107 [-]m-1
=1.09678 x 107 [-]m-1 102.70x 105 m-1 =1.09678 x 107 [-]m-1
-]m-1 =[-]
-] 0.936=1-
=1-0.936=0.064
=
=16 n2 =
=4 Answer
=4 Answer(b) Indicate the name of the series to which this photon belongs.
Ans. Since the electron falls from n=4 to n=1, therefore, the name of the series is Lyman series
(d) If the electron will fall from higher orbit to n=2, then calculate the wave number of the photon emitted. Why this energy difference is so small as compared to that in part (a)?
Solution:
n1 =2 ; n2=4
=? Formula: =1.09678 x 107 [-]m-1
=1.09678 x 107 [-]m-1=1.09678 x 107 [
-
]m-1=1.09678 x 107 [-
]m-1=1.09678 x 107 [
]m-1=0.20565x 107 m-1=0.20565x 105m-1 AnswerQ25. (a) What is de Broglie’s wavelength of an electron in meters traveling at half a speed of light ?
m =9.109 x 10-31 kg , c=3x 108 ms-1
Solution:
M=9.109 x 10-31 kg : v=
x3x 108 ms-1 =1.5x 108 ms-1
x3x 108 ms-1 =1.5x 108 ms-1 h =6.624 x 10-34 js =6.624 x 10-34 kg m2 s-1
Formula: =
==
=0.485 x 10-11 m=0.485 x 10-12 m=0.485 x 10-12 x 1010 =0.485 x 10-2 =0.0485 Answer(b) Convert the mass of electron into grams and velocity of light into cms-1 , and then calculate the wavelength of an electron in cm.
Solution:
m=9.109 x 10-31 kg =9.109 x 10-31 x 103 g =9.109 x 10-28 g
c =3 x 10-8 ms-1 =3 x 10-8 x 102 cms-1 =3 x 108 cms-1
=4.85 x 10-12 m=4.85 x 10-12 x102 cm =4.85 x 10-10 cm =0.048 x 10-8 cm Answer
(c) Convert the wavelength of electron from meters to
i) nm ii) iii) pm
iii) pmSolution:
i) =4.85 x 10-12 m=4.85 x 10-12 x109 cm =4.85 x 10-3 nm=0.048nm
=4.85 x 10-12 m=4.85 x 10-12 x109 cm =4.85 x 10-3 nm=0.048nmii) =4.85 x 10-12 m=4.85 x 10-12 x1010=4.85 x 10-2 =0.0485 Ans.
=4.85 x 10-12 m=4.85 x 10-12 x1010=4.85 x 10-2 =0.0485 Ans.iii) =4.85 x 10-12 m=4.85 x 10-12 x10-12 pm =4.485 x 10-24 pm Ans.
=4.85 x 10-12 m=4.85 x 10-12 x10-12 pm =4.485 x 10-24 pm Ans.
20 comments:
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Hi, I was wondering if you could give the answer to Question 12-b? It's not included in the post.
And great post btw :)
thnxxxx
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ThAnks For sUch A niCE nOtES
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Why cathod rays make a shadow of opaque object placed in thier path are they emit light
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give reason ...........cathode rays have reducing effect..................................plz tell me
The answer to question 13 is not included in notes plaza add it
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Tell me why anode rays are affected by the nature of gas
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BUT ALSO ADD LONG Qs
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Answer of question no 9
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