Chapter 12 Atoms Solutions
Question - 11 : - The gravitational attractionbetween electron and proton in a hydrogen atom is weaker than the coulombattraction by a factor of about 10−40. An alternative way of lookingat this fact is to estimate the radius of the first Bohr orbit of a hydrogenatom if the electron and proton were bound by gravitational attraction. Youwill find the answer interesting.
Answer - 11 : -
Radius of the first Bohr orbit isgiven by the relation,
Where,
∈0 = Permittivity of free space
h =Planck’s constant = 6.63 × 10−34 Js
me =Mass of an electron = 9.1 × 10−31 kg
e =Charge of an electron = 1.9 × 10−19 C
mp =Mass of a proton = 1.67 × 10−27 kg
r =Distance between the electron and the proton
Coulombattraction between an electron and a proton is given as:
Gravitational force of attractionbetween an electron and a proton is given as:
Where,
G = Gravitational constant = 6.67× 10−11 N m2/kg2
If the electrostatic (Coulomb)force and the gravitational force between an electron and a proton are equal,then we can write:
∴FG = FC
Putting the value of equation (4)in equation (1), we get:
It is known that the universe is156 billion light years wide or 1.5 × 1027 m wide. Hence, wecan conclude that the radius of the first Bohr orbit is much greater than theestimated size of the whole universe.
Question - 12 : - Obtain an expression for thefrequency of radiation emitted when a hydrogen atom de-excites from level n tolevel (n−1). For large n, show that this frequency equalsthe classical frequency of revolution of the electron in the orbit.
Answer - 12 : -
It is given that a hydrogen atomde-excites from an upper level (n) to a lower level (n−1).
We have the relation for energy (E1)of radiation at level n as:
Now, the relation for energy (E2) ofradiation at level (n − 1) is
givenas:
Energy (E) released as aresult of de-excitation:
E = E2−E1
hν = E2 − E1 …(iii)
Where,
ν = Frequency of radiationemitted
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 anelectron is given as:
Where,
Velocity of the electron inthe nth orbit is given as:
v =
And, radius of the nth orbitis given as:
r = 
Putting the values of equations(vi) and (vii) in equation (v), we get:
Hence, the frequency of radiationemitted by the hydrogen atom is equal to its classical orbital frequency.
Question - 13 : - Classically, an electron can bein any orbit around the nucleus of an atom. Then what determines the typicalatomic size? Why is an atom not, say, thousand times bigger than its typicalsize? The question had greatly puzzled Bohr before he arrived at his famousmodel of the atom that you have learnt in the text. To simulate what he mightwell have done before his discovery, let us play as follows with the basicconstants of nature and see if we can get a quantity with the dimensions oflength that is roughly equal to the known size of an atom (~ 10−10 m).
(a) Constructa quantity with the dimensions of length from the fundamental constants e, me,and c. Determine its numerical value.
(b) You will find that thelength obtained in (a) is many orders of magnitude smaller than the atomicdimensions. Further, it involves c. But energies of atoms aremostly in non-relativistic domain where c is not expected toplay any role. This is what may have suggested Bohr to discard c andlook for ‘something else’ to get the right atomic size. Now, the Planck’sconstant h had already made its appearance elsewhere. Bohr’sgreat insight lay in recognising that h, me,and e will yield the right atomic size. Construct a quantitywith the dimension of length from h, me,and e and confirm that its numerical value has indeed thecorrect order of magnitude.
Answer - 13 : -
(a) Chargeon an electron, e = 1.6 × 10−19 C
Mass of an electron, me =9.1 × 10−31 kg
Speed of light, c =3 ×108 m/s
Letus take a quantity involving the given quantities as 
Where,
∈0 = Permittivity of free space
And, 
The numerical value of the takenquantity will be:
Hence, the numerical value of thetaken quantity is much smaller than the typical size of an atom.
(b) Chargeon an electron, e = 1.6 × 10−19 C
Mass of an electron, me =9.1 × 10−31 kg
Planck’s constant, h =6.63 ×10−34 Js
Letus take a quantity involving the given quantities as 
Where,
∈0 = Permittivity of free space
And,
The numerical value of the takenquantity will be:
Hence, the value of the quantitytaken is of the order of the atomic size.
The total energy of an electronin the first excited state of the hydrogen atom is about −3.4 eV.
(a) What isthe kinetic energy of the electron in this state?
(b) Whatis the potential energy of the electron in this state?
(c) Which ofthe answers above would change if the choice of the zero of potential energy ischanged?
Answer - 14 : -
(a) Totalenergy of the electron, E = −3.4 eV
Kinetic energy of the electron isequal to the negative of the total energy.
K = −E
= − (− 3.4) = +3.4 eV
Hence, the kinetic energy of theelectron in the given state is +3.4 eV.
(b) Potentialenergy (U) of the electron is equal to the negative of twice of itskinetic energy.
U =−2 K
= − 2 × 3.4 = − 6.8 eV
Hence, the potential energy ofthe electron in the given state is − 6.8 eV.
(c) Thepotential energy of a system depends on the reference point taken. Here, thepotential energy of the reference point is taken as zero. If the referencepoint is changed, then the value of the potential energy of the system alsochanges. Since total energy is the sum of kinetic and potential energies, totalenergy of the system will also change.
Question - 15 : - If Bohr’s quantisation postulate(angular momentum = nh/2π) is a basic law of nature, it should beequally valid for the case of planetary motion also. Why then do we never speakof quantisation of orbits of planets around the sun?
Answer - 15 : -
We never speak of quantization oforbits of planets around the Sun because the angular momentum associated withplanetary 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 1070h.This leads to a very high value of quantum levels n of theorder of 1070. For large values of n, successiveenergies and angular momenta are relatively very small. Hence, the quantumlevels for planetary motion are considered continuous.
Question - 16 : - Obtain the first Bohr’s radiusand the ground state energy of a muonic hydrogen atom [i.e.,an atom in which a negatively charged muon (μ−) of mass about 207me orbitsaround a proton].
Answer - 16 : - Mass of a negatively charged muon, 
According to Bohr’s model,
Bohrradius, 
And, energy of a ground state electronic hydrogen atom,
We have the value of the first Bohr orbit, 
Let rμ bethe radius of muonic hydrogen atom.
At equilibrium, we can write therelation as:
Hence, the value of the firstBohr radius of a muonic hydrogen atom is
2.56 × 10−13 m.
We have,
Ee= − 13.6eV
Take the ratio of these energiesas:
Hence, the ground state energy ofa muonic hydrogen atom is −2.81 keV.