Chapter 12 Atoms Solutions
Question - 1 : - Choose the correct alternativefrom the clues given at the end of the each statement:
(a) The size of the atom in Thomson’smodel is .......... the atomic size in Rutherford’s model. (much greaterthan/no different from/much less than.)
(b) In the ground state of ..........electrons are in stable equilibrium, while in .......... electrons alwaysexperience a net force.
(Thomson’smodel/ Rutherford’s model.)
(c) A classical atombased on .......... is doomed to collapse.
(Thomson’smodel/ Rutherford’s model.)
(d) An atom has a nearly continuousmass distribution in a .......... but has a highly non-uniform massdistribution in ..........
(Thomson’smodel/ Rutherford’s model.)
(e) The positively charged part ofthe atom possesses most of the mass in .......... (Rutherford’s model/both themodels.)
Answer - 1 : -
(a) Thesizes of the atoms taken in Thomson’s model and Rutherford’s model have thesame order of magnitude.
(b) In the ground state of Thomson’smodel, the electrons are in stable equilibrium. However, in Rutherford’s model,the electrons always experience a net force.
(c) A classical atombased on Rutherford’s model is doomed to collapse.
(d) An atom has a nearly continuous massdistribution in Thomson’s model, but has a highly non-uniform mass distributionin Rutherford’s model.
(e) The positively charged part of the atompossesses most of the mass in both the models.
Question - 2 : - Suppose you are given a chance torepeat the alpha-particle scattering experiment using a thin sheet of solidhydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below14 K.) What results do you expect?.
Answer - 2 : -
In the alpha-particle scatteringexperiment, 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 massof hydrogen (1.67 × 10−27 kg) isless 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 solidhydrogen is used in the α-particle scattering experiment.
Question - 3 : - In the alpha-particle scatteringexperiment, 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 massof hydrogen (1.67 × 10−27 kg) isless 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 solidhydrogen is used in the α-particle scattering experiment..
Answer - 3 : -
Rydberg’s formula is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
(n1 and n2 are integers)
Theshortest wavelength present in the Paschen series of the spectral lines isgiven for values n1 = 3 and n2 = ∞.
Question - 4 : - A difference of 2.3 eV separatestwo energy levels in an atom. What is the frequency of radiation emitted whenthe atom makes a transition from the upper level to the lower level?
Answer - 4 : -
Separation of two energy levelsin an atom,
E = 2.3 eV
= 2.3 ×1.6 × 10−19
= 3.68 ×10−19 J
Let ν bethe frequency of radiation emitted when the atom transits from the upper levelto the lower level.
We havethe relation for energy as:
E = hv
Where,
h =Planck’s constant
Hence, the frequency of theradiation is 5.6 × 1014 Hz.
The ground state energy ofhydrogen atom is −13.6 eV. What are the kinetic and potential energies of theelectron in this state?
Answer - 5 : -
Ground state energy of hydrogenatom, E = − 13.6 eV
This isthe total energy of a hydrogen atom. Kinetic energy is equal to the negative ofthe total energy.
Kineticenergy = − E = − (− 13.6) =13.6 eV
Potentialenergy is equal to the negative of two times of kinetic energy.
Potentialenergy = − 2 × (13.6) = − 27 .2eV
Question - 6 : - A hydrogen atom initially in theground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelengthand frequency of the photon.
Answer - 6 : -
For ground level, n1 = 1
Let E1 be the energy of this level. It is known that E1 is related with n1 as:
The atom is excited to a higherlevel, n2 = 4.
Let E2 be the energy of this level.
The amount of energy absorbed bythe photon is given as:
E = E2 − E1
For a photon of wavelengthλ, the expression of energy is written as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
And, frequency of a photon isgiven by the relation,
Hence, the wavelength of thephoton is 97 nm while the frequency is 3.1 × 1015 Hz.
Question - 7 : - (a) Using the Bohr’s modelcalculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculatethe orbital period in each of these levels.
Answer - 7 : -
(a) Let ν1 be the orbital speed of the electron in ahydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
Where,
e = 1.6 × 10−19 C
∈0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
h = Planck’s constant = 6.62 × 10−34 Js
For level n2 = 2, we can write the relation for thecorresponding orbital speed as:
And, for n3 = 3, we can write the relation for thecorresponding orbital speed as:
Hence, the speed of the electronin a hydrogen atom in n = 1,n=2, and n=3 is 2.18 × 106 m/s,1.09 × 106 m/s, 7.27 × 105 m/s respectively.
(b) Let T1 be the orbitalperiod of the electron when it is in level n1 = 1.
Orbitalperiod is related to orbital speed as:
Where,
r1 =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 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
For level n2 = 2, we can write the period as:
Where,
r2 =Radius of the electron in n2 = 2
And, for level n3 = 3, we can write the period as:
Where,
r3 =Radius of the electron in n3 = 3
Hence, the orbital period in eachof these levels is 1.52 × 10−16 s,1.22 × 10−15 s, and 4.12 × 10−15 s respectively.
Question - 8 : - The radius of the innermostelectron orbit of a hydrogen atom is 5.3 ×10−11 m.What are the radii of the n = 2and n =3 orbits?
Answer - 8 : -
The radius of the innermost orbitof a hydrogen atom, r1 = 5.3 × 10−11 m.
Let r2 be the radius of the orbit at n = 2. It is related to the radius ofthe innermost orbit as:
For n =3, we can write the corresponding electron radius as:
Hence, the radii of an electronfor n = 2 and n = 3 orbits are 2.12 × 10−10 m and 4.77 × 10−10 m respectively.
Question - 9 : - A 12.5 eV electron beam is usedto bombard gaseous hydrogen at room temperature. What series of wavelengthswill be emitted?
Answer - 9 : -
It is given that the energy ofthe electron beam used to bombard gaseous hydrogen at room temperature is 12.5eV. Also, the energy of the gaseous hydrogen in its ground state at roomtemperature is −13.6 eV.
When gaseous hydrogen isbombarded 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 toorbit level (n) as:
For n = 3,
This energy is approximatelyequal to the energy of gaseous hydrogen. It can be concluded that the electronhas jumped from n = 1 to n = 3 level.
During its de-excitation, theelectrons can jump from n = 3 to n = 1directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wavenumber for Lyman series as:
Where,
Ry =Rydberg constant = 1.097 × 107 m−1
λ= Wavelength ofradiation emitted by the transition of the electron
For n = 3, wecan obtain λas:
If the electron jumps from n =2 to n = 1, then the wavelength of the radiation is given as:
If the transition takes placefrom n = 3 to n = 2, then the wavelength of the radiation is given as:
This radiation corresponds to theBalmer series of the hydrogen spectrum.
Hence, in Lyman series, twowavelengths 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.
Question - 10 : - Answer the following questions,which help you understand the difference between Thomson’s model and Rutherford’smodel better.
(a) Is theaverage angle of deflection of α-particles by a thin gold foilpredicted by Thomson’s model much less, about the same, or much greater thanthat predicted by Rutherford’s model?
(b) Is theprobability of backward scattering (i.e., scattering of α-particlesat angles greater than 90°) predicted by Thomson’s model much less, about thesame, or much greater than that predicted by Rutherford’s model?
(c) Keepingother factors fixed, it is found experimentally that for small thickness t,the number of α-particles scattered at moderate angles isproportional to t. What clue does this linear dependence on t provide?
(d) Inwhich model is it completely wrong to ignore multiple scattering for thecalculation of average angle of scattering of α-particles by a thinfoil?
Answer - 10 : -
(a) about thesame
The average angle of deflectionof α-particles by a thin gold foil predicted by Thomson’s model isabout the same size as predicted by Rutherford’s model. This is because theaverage angle was taken in both models.
(b) much less
The probability of scatteringof α-particles at angles greater than 90° predicted by Thomson’smodel is much less than that predicted by Rutherford’s model.
(c) Scatteringis mainly due to single collisions. The chances of a single collision increaseslinearly with the number of target atoms. Since the number of target atomsincrease with an increase in thickness, the collision probability dependslinearly on the thickness of the target.
(d) Thomson’smodel
It is wrong to ignore multiplescattering in Thomson’s model for the calculation of average angle ofscattering of α−particles by a thin foil. This is because a singlecollision causes very little deflection in this model. Hence, the observedaverage scattering angle can be explained only by considering multiplescattering.