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#### Page No 435:

#### Question 12.1:

Choose the correct alternative from the clues given at the 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)** The sizes of the atoms taken in Thomson’s model and
Rutherford’s model have the same order of magnitude.

**(b) ** In the ground state of Thomson’s model, the
electrons are in stable equilibrium. However, in Rutherford’s
model, the electrons always experience a net force.

**(c)** A *classical
*atom based on Rutherford’s model is doomed to
collapse.

**(d)** An atom has a nearly continuous mass distribution in
Thomson’s model, but has a highly non-uniform mass distribution
in Rutherford’s model.

**(e)** The positively charged part of the atom possesses most of
the mass in both the models.

#### Page No 435:

#### Question 12.2:

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:

In the alpha-particle
scattering experiment, if a thin sheet of solid hydrogen is used in
place of a gold foil, then the scattering angle would not be large
enough. This is because the mass of hydrogen (1.67 × 10^{−27
}kg) is less than the mass of incident α−particles
(6.64 × 10^{−27 }kg). Thus, the mass of the
scattering particle is more than the target nucleus (hydrogen). As a
result, the α−particles
would not bounce back if solid hydrogen is used in the α-particle
scattering experiment.

#### Page No 436:

#### Question 12.3:

What is the shortest wavelength present in the Paschen series of spectral lines?

#### Answer:

Rydberg’s formula is given as:

Where,

*h* = Planck’s
constant = 6.6 × 10^{−34} Js

*c *= Speed of
light = 3 × 10^{8 }m/s

*(n*_{1}
and *n*_{2} are integers)

The shortest wavelength
present in the Paschen series of the spectral lines is given for
values *n*_{1} = 3 and *n*_{2} = ∞.

#### Page No 436:

#### Question 12.4:

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

#### Answer:

Separation of two energy levels in an atom,

*E* = 2.3 eV

= 2.3 × 1.6 ×
10^{−19}

= 3.68 × 10^{−19}
J

Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level.

We have the relation for energy as:

*E* = *hv*

Where,

*h *= Planck’s
constant

Hence, the frequency of
the radiation is 5.6 × 10^{14} Hz.

#### Page No 436:

#### Question 12.5:

The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?

#### Answer:

Ground state energy of
hydrogen atom, *E* = − 13.6 eV

This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.

Kinetic energy = −
*E *= − (− 13.6) = 13.6 eV

Potential energy is equal to the negative of two times of kinetic energy.

Potential energy* =*
− 2 × (13.6) = − 27 .2 eV

#### Page No 436:

#### Question 12.6:

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 the photon.

#### Answer:

For ground level, *n*_{1}
= 1

Let *E*_{1}
be the energy of this level. It is known that *E*_{1} is
related with *n*_{1} as:

The atom is excited to
a higher level, *n*_{2} = 4.

Let *E*_{2}
be the energy of this level.

The amount of energy absorbed by the photon is given as:

*E *= *E*_{2}
− *E*_{1}

For a photon of
wavelength*λ*, the
expression of energy is written as:

Where,

*h* = Planck’s
constant = 6.6 × 10^{−34} Js

*c* = Speed of
light = 3 × 10^{8} m/s

And, frequency of a photon is given by the relation,

Hence, the wavelength
of the photon is 97 nm while the frequency is 3.1 × 10^{15}
Hz.

#### Page No 436:

#### Question 12.7:

(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)** Let *ν*_{1} be the orbital speed of the
electron in a hydrogen atom in the ground state level, *n*_{1}
= 1. For charge (*e) *of an electron, *ν*_{1 }is
given by the relation,

Where,

*e
*= 1.6 × 10^{−19} C

∈_{0}
= Permittivity of free space = 8.85 × 10^{−12}
N^{−1} C^{2} m^{−2}

*h*
= Planck’s constant = 6.62 × 10^{−34} Js

For
level *n*_{2} = 2, we can write the relation for the
corresponding orbital speed as:

And,
for *n*_{3} = 3, we can write the relation for the
corresponding orbital speed as:

Hence,
the speed of the electron in a hydrogen atom in *n* = 1, n=2,
and n=3 is 2.18 × 10^{6} m/s, 1.09 × 10^{6}
m/s, 7.27 × 10^{5} m/s respectively.

**(b)** Let *T*_{1}
be the orbital period of the electron when it is in level *n*_{1}
= 1.

Orbital period is related to orbital speed as:

Where,

*r*_{1}
= Radius of the orbit

*h*
= Planck’s constant = 6.62 × 10^{−34} Js

*e*
= Charge on an electron = 1.6 × 10^{−19} C

∈_{0
}= Permittivity of free space = 8.85 × 10^{−12}
N^{−1} C^{2} m^{−2}

*m*
= Mass of an electron = 9.1 × 10^{−31} kg

For
level *n*_{2} = 2, we can write the period as:

Where,

*r*_{2}
= Radius of the electron in *n*_{2} = 2

And,
for level *n*_{3} = 3, we can write the period as:

Where,

*r*_{3}
= Radius of the electron in *n*_{3} = 3

Hence,
the orbital period in each of these levels is 1.52 × 10^{−16}
s, 1.22 × 10^{−15} s, and 4.12 × 10^{−15}
s respectively.

#### Page No 436:

#### Question 12.8:

The radius of the
innermost electron orbit of a hydrogen atom is 5.3 ×10^{−11}
m. What are the radii of the *n *= 2 and *n *=3 orbits?

#### Answer:

The radius of the
innermost orbit of a hydrogen atom, *r*_{1} = 5.3 ×
10^{−11} m.

Let *r*_{2}
be the radius of the orbit at *n* = 2. It is related to the
radius of the innermost orbit as:

For *n* = 3, we
can write the corresponding electron radius as:

Hence, the radii of an
electron for *n* = 2 and *n* = 3 orbits are 2.12 ×
10^{−10} m and 4.77 × 10^{−10} m
respectively.

#### Page No 436:

#### Question 12.9:

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

#### Answer:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is
related to orbit level (*n)* as:

For *n* = 3,

This energy is
approximately equal to the energy of gaseous hydrogen. It can be
concluded that the electron has jumped from *n* = 1 to *n*
= 3 level.

During its
de-excitation, the electrons can jump from *n* = 3 to *n* =
1 directly, which forms a line of the Lyman series of the hydrogen
spectrum.

We have the relation for wave number for Lyman series as:

Where,

*R*_{y} =
Rydberg constant = 1.097 × 10^{7} m^{−1}

λ*=*
Wavelength of radiation emitted by the transition of the electron

For *n* = 3, we
can obtain *λ*as:

If the electron jumps
from *n* = 2 to *n* = 1, then the wavelength of the
radiation is given as:

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

#### Page No 436:

#### Question 12.10:

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 ×
10^{11} m with orbital speed 3 × 10^{4} m/s.
(Mass of earth = 6.0 × 10^{24} kg.)

#### Answer:

Radius of the orbit of
the Earth around the Sun, *r* = 1.5 × 10^{11} m

Orbital speed of the
Earth, *ν* = 3 × 10^{4} m/s

Mass of the Earth, *m*
= 6.0 × 10^{24} kg

According to Bohr’s model, angular momentum is quantized and given as:

Where,

*h* = Planck’s
constant = 6.62 × 10^{−34} Js

*n* = Quantum
number

Hence, the quanta
number that characterizes the Earth’ revolution is 2.6 ×
10^{74}.

#### Page No 436:

#### Question 12.11:

Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

**(a) ** Is the average angle of deflection of *α***-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 *α*-particles
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 *α*-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
*α*-particles by a thin
foil?

#### Answer:

**(a) ** about the
same

The
average angle of deflection of *α*-particles
by a thin gold foil predicted by Thomson’s model is about the
same size as predicted by Rutherford’s model. This is because
the average angle was taken in both models.

**(b) **much less

The
probability of scattering of *α*-particles
at angles greater than 90°
predicted by Thomson’s model is much less than that predicted
by Rutherford’s model.

**(c) **Scattering is mainly due to single collisions. The chances
of a single collision increases linearly with the number of target
atoms. Since the number of target atoms increase with an increase in
thickness, the collision probability depends linearly on the
thickness of the target.

**(d) **Thomson’s
model

It
is wrong to ignore multiple scattering in Thomson’s model for
the calculation of average angle of scattering of *α*−particles
by a thin foil. This is because a single collision causes very little
deflection in this model. Hence, the observed average scattering
angle can be explained only by considering multiple scattering.

#### Page No 436:

#### Question 12.12:

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:

Radius of the first Bohr orbit is given by the relation,

Where,

∈_{0} =
Permittivity of free space

*h* = Planck’s
constant = 6.63 × 10^{−34} Js

*m*_{e} =
Mass of an electron = 9.1 × 10^{−31} kg

*e* = Charge of an
electron = 1.9 × 10^{−19} C

*m*_{p} =
Mass of a proton = 1.67 × 10^{−27} kg

*r* = Distance
between the electron and the proton

Coulomb attraction between an electron and a proton is given as:

Gravitational force of attraction between an electron and a proton is given as:

Where,

G = Gravitational
constant = 6.67 × 10^{−11} N m^{2}/kg^{2}

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

∴*F*_{G}
= *F*_{C}

Putting the value of equation (4) in equation (1), we get:

It is known that the
universe is 156 billion light years wide or 1.5 × 10^{27 }m
wide. Hence, we can conclude that the radius of the first Bohr orbit
is much greater than the estimated size of the whole universe.

#### Page No 436:

#### Question 12.13:

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:

It is given that a
hydrogen atom de-excites from an upper level (*n)* to a lower
level (*n*−1).

We have the relation
for energy (*E*_{1}) of radiation at level *n *as:

Now, the relation for
energy (*E*_{2}) of radiation at level (*n *−
1) is givenas:

Energy (*E*)
released as a result of de-excitation:

*E* = *E*_{2}−*E*_{1}

*h**ν*
= *E*_{2} − *E*_{1} …
(iii)

Where,

ν = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii), we get:

For large *n,* we
can write

Classical relation of frequency of revolution of an electron is given as:

Where,

Velocity of the
electron in the *n*^{th} orbit is given as:

*v* =

And, radius of the *n*^{th}
orbit is given as:

*r* =

Putting the values of equations (vi) and (vii) in equation (v), we get:

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

#### Page No 437:

#### Question 12.14:

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^{−10
}m).

**(a) **Construct a quantity with the dimensions of length from
the fundamental constants *e*, *m*_{e}, 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 *h*, *m*_{e}, and *e *will yield
the right atomic size. Construct a quantity with the dimension of
length from *h*, *m*_{e}, and *e *and
confirm that its numerical value has indeed the correct order of
magnitude.

#### Answer:

**(a)** Charge on an
electron, *e* = 1.6 × 10^{−19 }C

Mass
of an electron,* m*_{e} = 9.1 × 10^{−31}
kg

Speed
of light, *c* = 3 ×10^{8} m/s

Let us take a quantity involving the given quantities as

Where,

∈_{0}
= Permittivity of free space

And,

The numerical value of the taken quantity will be:

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

**(b)** Charge on an electron, *e* = 1.6 × 10^{−19
}C

Mass
of an electron,* m*_{e} = 9.1 × 10^{−31}
kg

Planck’s
constant, *h* = 6.63 ×10^{−34} Js

Let us take a quantity involving the given quantities as

Where,

∈_{0}
= Permittivity of free space

And,

The numerical value of the taken quantity will be:

Hence, the value of the quantity taken is of the order of the atomic size.

#### Page No 437:

#### Question 12.15:

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 changed?

#### Answer:

**(a) **Total energy
of the electron, *E* = −3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy.

*K*
= −*E*

= − (− 3.4) = +3.4 eV

Hence, the kinetic energy of the electron in the given state is +3.4 eV.

**(b) **Potential energy (*U*) of the electron is equal to
the negative of twice of its kinetic energy.

*U*
= −2 *K*

= − 2 × 3.4 = − 6.8 eV

Hence, the potential energy of the electron in the given state is − 6.8 eV.

**(c) **The potential energy of a system depends on the reference
point taken. Here, the potential energy of the reference point is
taken as zero. If the reference point is changed, then the value of
the potential energy of the system also changes. Since total energy
is the sum of kinetic and potential energies, total energy of the
system will also change.

#### Page No 437:

#### Question 12.16:

If Bohr’s
quantisation postulate (angular momentum = *nh/*2π)
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:

We never speak of
quantization of orbits of planets around the Sun because the angular
momentum associated with planetary motion is largely relative to the
value of Planck’s constant* (h)*. The angular momentum of
the Earth in its orbit is of the order of 10^{70}*h*.
This leads to a very high value of quantum levels *n *of the
order of 10^{70}. For large values of *n*, successive
energies and angular momenta are relatively very small. Hence, the
quantum levels for planetary motion are considered continuous.

#### Page No 437:

#### Question 12.17:

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 207m_{e} orbits around a proton].

#### Answer:

Mass of a negatively charged muon,

According to Bohr’s model,

Bohr radius,

And, energy of a ground
state *electronic* *hydrogen* *atom*,

We have the value of the first Bohr orbit,

Let *r*_{μ}
be the radius of *muonic hydrogen atom*.

At equilibrium, we can write the relation as:

Hence, the value of the
first Bohr radius of a *muonic hydrogen atom *is

2.56 × 10^{−13}
m.

We have,

*E*_{e}=
− 13.6 eV

Take the ratio of these energies as:

Hence, the ground state
energy of a *muonic hydrogen atom* is −2.81 keV.

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