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 Bihar Board Class 12 Physics Solutions Chapter 12 Atoms

Bihar Board Class 12 Physics Atoms Textbook Questions and Answers

Question 1.
Choose the correct alternative from clues given at end of the each statement :
(a) The size of the atom in Thomson’s model is ………….. the atomic size in Rutherford’s model, (much greater than/no different from/ much less than.)
(b) In the ground state of ………….. electrons are in stable equilibrium, while in electrons always experience a net force.
(Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on ………….. is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a ………….. but has a highly non-uniform mass distribution in ………… (Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in ………….. (Rutherford’s model/both the models.)
Answer:
(a) No different from.
(b) Thomson’s model, Rutherford’s model.
(c) Rutherford’s model.
(d) Thomson’s model, Rutherford’s model.
(e) Both the models.

Question 2.
Suppose you are given a chance to repeat the alpha- particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
Answer:
The nucleus of a hydrogen atom is a proton. The mass of it is 1.67 x 10-27 kg, whereas the mass of an incident α-particle is 6.64 x 10-27 kg. Because the scattering particle is more massive than the target nuclei (proton), the α-particle won’t bounce back in even in a head-on collision. It is similar to a football colliding with a tenis ball at rest. Thus, there would be no large-angle scattering.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
We know that the wave number of the emitted radiation is given by

For Paschen series, n1= 3, n2 = ∞ = for shortest wavelength. If λmin be the minimum wavelength in this series of spectral lines, then –

Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Answer:
Here, AE = 2.3 eV = 2.3 x 1.6 x 10-19 J.
h = 6.62 x 10-34JS.
v = frequency of radiation emitted = ?
Using the relation,

Question 5.
The ground state energy of hydrogen atom is -13.6 eV. What is the kinetic and potential energies of the electron in this state?
Answer:
We know that the K.E. of the electron in the nth orbit of hydrogen atom is given by
T = K.E = 12 m vn2 = 2π2k2me4n2 h2 where K = 14πε0
for ground state, n = 1
∴ T = K.E = 2π2k2me4 h2 …………(1)
Also we know that the P.E. of the electron in the nth orbit of hydrogen atom is given by
V = P.E = – K e2rn = – k2⋅4π2me4nh2
∴ V = P.E. in ground state is given by
V = P.E = – 4π2K2 me4 h2 ………(2)

If E be the total energy in the ground state, then
E = K.E. + P.E. = T + V = −2π2k2me4h2
= -T. ………..(3)
Here, E = Total energy in ground state = -13,6 eV. T = ?, V = ?
∴ From (3), T = – E = (- 13.6) = 13.6 eV.
and V = E – T = -T – T = – 2T
= – 2 x 13.6 = – 27.2 eV.
K.E. = 13.6 eV, P.E. = – 27.2 eV

Question 6.
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Answer:
We know that the energy En of the hydrogen atom in the nth stationary state is given by :
En = −13.6n2
For ground state, n = 1,
∴ En = E4
and for n = 4 level, En = En.
h = 6.62 x 10-34 Js.
Let λ and v be the wavelength and frequency of the photon absorbed by the hydrogen atom = ?
If E = hv be the energy absorbed, then
hv = E4 – E1 ……..(1)
Now E4 = −13.642=−13.616 = – 0.85 eV.
and Et = -13.6eV. ……(2)
∴ From (1) and (2), we get

Question 7.
(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1,2, and 3 levels,
(b) Calculate the orbital period in each of these levels.
Answer:
(a) According to Bohr’s model, we know that the radius of nth orbit and speed of electron in the nth orbit are given by :

Let v1, v2 and v3 be speeds of electron in the orbits n = 1,2 and 3.

(b) Let T1, T2 and T3 be the orbital time period in level no. 1,2 and 3 resepectively.
Now we know that time period in nth orbit is given by :
Tn = 2πrnvn
Using eq (1) and (2), we get

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 1011 m. What are the radii of the n = 2 and n = 3 orbits?
Answer:
We know that the radius of the innermost orbit of electron in the hydrogen atom is given by :
rn = 4πε0n2h24π2me2 = Knn
where K = 4πε0 h24π2me2
where h = 6.62 x 10-34 Js, m = 9.1 x 10-31 kg, e = 1.6 x 10-19C.
for n = 1, rn = r1
r1 = 5.3 x 10-11m = K.
for n = 2,
r2 = KZ2 = 4r1 = 4 x 5.3 x 10-11 = 21.2 x 10-11m
= 2.12 x 10-10 m
for n = 3, r2 = K2 32 = 9 r2 = 9 x 5.3 x 10-11 H = 47.7 x 10-11 m
= 4.77 x 10-10 m.

Question 9.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:
Here, E = energy of electron beam = 12.5 eV
= 12.5 x 1.6 x 10-19J.
h = 6.62 x 10-34JS
C = 3 x 108 ms-1
Let λ = wavelength of radiation emitted due to bombardment of electron beam = ?
Using the relation,
E = hCλ , we get

This wavelength falls in the range of Lyman series (912 Å to 1216 Å), thus we conclude that Lyman series of wavelength 993 Å is emitted.

Question 10.
In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 x 1011 m with orbital speed 3 x 104 m/s. (Mass of earth = 6.0 x 1024 kg.)
Answer:
Here, r = radius of orbit of earth around the sun = 1.5 x 1011
m = mass of earth = 6.0 x 1024 kg.
v = orbital speed of earth = 3 x 104 ms-1,
h = 6.62 x 10-34 JS.
n = quantum number of the orbit of earth around the sun in which it revolves around the sun = ?
Using the relation,

The quantum number for the satellite motion is extremely large. In fact for such large quantum numbers, the results of quantisation conditions tends to those of classical physics i.e., the difference between the two successive energy or angular momentum levels is very small and the levels may be considered continuous.

Additional Exercises

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’ model better.
(a) Is the average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of expat-tides at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of a-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a- particles by a thin foil?
Answer:
(a) The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same as that predicted by Rutherford’s atom model. This is because we are talking of average angle of deflection.
(b) The probability of backward scattering predicted by Thomson’s model is much less than that predicted by Rutherford’s model. This is because there is no such massive central core called the nucleus as in Rutherford’s model.
(c) This linear dependence on suggests that scattering is predominantly due to a single collision, because chance of a single collision increases with the number of target atoms which increases linearly with the thickness of the foil.
(d) In Thomson model, positive charge is uniformly distributed in the spherical atom. Therefore, a single collision causes very little deflection Thus average scattering angle can be explained only by considering multiple scattering in Thomson model.
On the other hand, in Rutherford’s model, most of the scattering comes from a single collision. So multiple scattering may be ignored.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10’40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
Let mp and me be the masses of proton and electron respectively lying at a distance r.
If F be the gravitational force between them, then
F = G mempr2 …………(1)
This force provides the necessary centrepetal force for uniform circular motion :

Also according to Bohr’s model, the radius of first orbit is given by
r1 = 4πε0 h2 mee2 …………(7)
From (6) and (7), we conclude that if the place of electrostatic force (=e24πε0r2) , We consider the atom bound by the gravitational force (= G = mpme/r2) then we should replace e24πε0 by G mp me. thus the radius of the first Bohr orbit in gravitional bound hydrogen atoms is given by

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – 1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The energy of an electron in the nth orbit of hydrogen atom is given by the relation,
E = −2π2me4k2n2h2 …………..(1)
Where K = 14πε0 is a constant.
Also the energy of an electron in the (n – 1) th orbit of hydrogen atom is given by the relation :
En – 1 = −2π2me4k2(n−1)2h2 ………..(2)
Thus if v be the frequency of the radiation emitted when the hydrogen atom de-excites from the nth level to (n – 1)th level, then

Also according to Bohr’s atom model the velocity of electron in the nth orbit is :
υn = nh2πmr
and the radius of the nth orbit is given by :
r = n2h24π2mke2
∴ the classical Frequency of revolution of electron is given by

From (4) and (5), we see that v = vc i.e., for large value of n, the classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when the hydrogen atom de-excites from level (n) to level (n – 1). This is called Bohr’s Correspondence principle.

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10-10m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.

(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non – relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that hr me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Here, the dimensional formula of e is [AT],
The dimensional formula of C = [LT– 1]
The dimensional formula of me = [M]
The dimensional formula of 14πε0 = [ML3 T-4 A2]

∴Comparing dimensions of M1 L1T and A in equation (2), we get
M, 0 = 1 + C or C = −C
L, 1 = 3 + b, ∴ b = 1 – 3 = – 2.
T,0 = -4 + 9 – b. ∴ a = 4 + b = 4 – 2 = 2
A, 0 = – 2 + A, ∴ A = 2.
Thus, eq (l), becomes :

(here e = 1.6 x 1019 C, m = 9.1 x 1031, C = 3 x 108 ms-1).
Which is very small as compared to the size of the atom (= 10-10 m).

(b) Thus from (a), we see that the magnitude of this quantity is many orders of magnitude smaller than the atomic dimensions.
Let the quantity x constructed from h, me and e given by

Comparing Powers of

is the required quantity having dimension of length. Its value is given by

Which is nearly equal to the size of the atom. Thus we confirm that its value is of the same order as the atomic size.

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about – 3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is charged?
Answer:
Here, total energy in first excited state (i.efor n = 2) of H – atom, E = – 3.4 eV.
K.E of electron = K he22r, Where K = 14πε0
And P.E = – Ze2r = -2 (kZe22r)
= – 2(K.E)
∴ E = Total energy in first excited state is given by
E = K.E. + P E, = K.E. + [- 2 (K.E.)] = – K,E,
Physics-XII (Part-II)
(a) ∴ K.E. = – E = – (- 3.4) = + 3.4 eV.
(b) P.E. = – 2 (K.E.) = – 2 x 3.4 = – 6.8 eV.
(c) If zero of the potential energy is changed, K.E. does not change and continues to be + 3.4 eV as it is independent of the choice of the zero of potential energy.
But the P.E. and total energy of the state would change with the choice of the zero of the potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?
Answer:
Applying Bohr’s quantisation postulate : we get
mvr = nh2π
or n = mv x 2πh …………(1)
For the motion of a planet (ray, earth) Let us take :
m = 6 x 1024 kg.
V = 3 x 104 kg.
r = 1.49 x 1011m.
h = 6.62 x 10-34 JS.
∴ From (1),
n = 6×1024×30000×1.49×1011×2×3.146.626×10−14
= 2.49 x 1074
i.e., n is very large. So the angular momentum associated with planetary motion are incomparably large relateive to h. For such a large value of n, the differences in the successive energies and angular moments of the quantised levels of the Bohr model are so small compared to the energies and angular momentum respectively of the levels that the levels may be considered continuous.

Question 17.
Obtain the first Bohr’s radius and the ground state energy of a ‘muonic hydrogen atom’ (i.e. an atom in which a negatively charged muon (μ–) of mass about 207 me orbits around a proton).
Answer:
A muonic hydrogen is the atom in which a negatively charged muon of mass about 20 % me revolves around a proton.
i.e., mu = 207 me = 207 x 9.1 x 10-31 kg.
Also according to Bohr’s model, r ∝ 1m
Let ru = first Bohr’s radius of muonic hydrogen atom.
rμre=memμ=1207 c
where re = radius of first orbit of the electron in hydrogen atom = 0.53 A =0.53 x 10-10 m.
∴ rμ = re207=0.53×10−10207 = 2.56 x 10-13 m-1
Let Eμ= ground state energy of a muonic hydrogen atom = ?
Also according to Bohr’s model,
E ∝ m.

Where Ee = ground state energy of electron in the hydrogen atom = – 13.6 eV.

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