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.)
Solution: No different From
(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.)
Solution: (i)Thomson's model ,(ii) Rutherford’s model
(c) A classical atom based on .......... is doomed to collapse.(Thomson’s model/ Rutherford’s model.)
Solution: 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.)
Solution : 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.)
Solution: Both the models
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?
Solution:
The nucleus of hydrogen contains only one proton. therefore hydrogen nuclei (protons) are much lighter than Alpha particles. when Alpha particle are incident on a thin sheet of solid hydrogen (in place of gold foil), the Alpha particle will not be scattered even on head on collision by solid hydrogen. it is because the scattering particle is more massive than that target particle. therefore we can't determine the size of hydrogen nucleus.
12.3 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?
Solution:
Here, $$E =2.3 eV = 2.3 \times 1.6 \times 10^{-19}J$$
$$E= 3.68 \times 10^{-19}J$$
$$E = hf$$
$$f = \frac{E}{h}$$
$$f= \frac{3.68 \times 10^{-19}}{6.62 \times 10^{-34}}$$
$$f = 5.6 \times 10^{14}Hz$$
12.4 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
Solution :
Total energy,
$$E =-13.6 eV$$
$$K. E = - E = - (-13.6 eV) = 13.6 eV$$
$$P. E = - 2 K. E = - 2 \times 13.6 = - 27.2 eV$$
12.5 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.
Solution :
The energy of an electron in the nth orbit of hydrogen atom is
$$E_{n} =\frac{-13.6}{n^{2}}$$
For n =1 (ground level)
$$E_{1} =\frac{-13.6}{(1)^{2}}$$
$$E_{1} = - 13.6 eV$$
For n =4
$$E_{4} =\frac{-13.6}{(4)^{2}}$$
$$E_{1} = - 0.85 eV$$
Energy of photon,
$$E = E_{4}-E_{1}$$
$$E = (-.85)-(-13.6)$$
$$E = 12.75 eV$$
$$E = 12.75 \times 1.6 \times 10^{-19}$$
$$E = 2.04 \times 10^{-18}J$$
Now
$$E = \frac{hc}{\lambda}$$
Or
$${\lambda} = \frac{hc}{E}$$
Wavelength of photon
$${\lambda} = \frac{6.62 \times 10^{-34}\times 3 \times 10^{8}}{2.04 \times 10^{-18}}$$
$${\lambda} = 9.7 \times 10^{-8}m$$
Frequency of photon
$$f = \frac{c}{\lambda}$$
$$f = \frac{3 \times 10^{8}}{9.7 \times 10^{-8}}$$
$$f = 3.1 \times 10^{15} Hz$$
12.6 (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.
Solution :
The radius of an electron in the nth orbit of hydrogen atom is
$$r_{n} = 0.53 \times 10^{-10} n^{2}$$
The speed of an electron in the nth orbit of hydrogen atom is
$$v_{n} = \frac{2.2 \times 10^{6}}{n}$$
For n =1,
$$v_{1} = \frac{2.2 \times 10^{6}}{1}$$
$$v_{1} = 2.2 \times 10^{6} m/s$$
For n =2,
$$v_{2} = \frac{2.2 \times 10^{6}}{2}$$
$$v_{2} = 1.1 \times 10^{6} m/s$$
For n =3,
$$v_{3} = \frac{2.2 \times 10^{6}}{3}$$
$$v_{3} = 0.733 \times 10^{6} m/s$$
$$v_{3} = 7.33 \times 10^{5} m/s$$
(b) The orbital period of an electron in the nth orbit of hydrogen atom is
$$T_{n} = \frac{2 \pi r_{n} }{v_{n} }$$
$$T_{n} = \frac{2 \pi \times 0.53 \times 10^{-10} n^{2}}{\frac{2.2 \times 10^{6}}{n} }$$
$$T_{n} = 1.52 \times 10^{-16}n^{3}$$
For n=1
$$T_{1} = 1.52 \times 10^{-16}.(1)^{3}$$
$$T_{1} = 1.52 \times 10^{-16}s$$
For n= 2
$$T_{2} = 1.52 \times 10^{-16}.(2)^{3}$$
$$T_{2} = 1.52 \times 10^{-16} \times 8$$
$$T_{2} = 12.16 \times 10^{-16}$$
$$T_{2} = 1.22 \times 10^{-15}s$$
For n= 3
$$T_{3} = 1.52 \times 10^{-16}.(3)^{3}$$
$$T_{3} = 1.52 \times 10^{-16} \times 27$$
$$T_{3} = 41.04 \times 10^{-16}s$$
$$T_{3} = 4.104 \times 10^{-15}s$$
12.7 The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11}m$. What are the radii of the n = 2 and n =3 orbits?
Solution :
$$r_{n} = 5.3 \times 10^{-11} n^{2}$$
$$r_{n} = 0.53 \times 10^{-10} n^{2}$$
For n=2,
$$r_{2} = 0.53 \times 10^{-10}. 2^{2}$$
$$r_{2} = 0.53 \times 10^{-10} \times 4$$
$$r_{2} = 2.12 \times 10^{-10}m$$
For n=3,
$$r_{3} = 0.53 \times 10^{-10}. 3^{2}$$
$$r_{3} = 0.53 \times 10^{-10} \times 9$$
$$r_{3} = 4.77 \times 10^{-10}m$$
12.8 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Solution:
The electron beam energy used at room temperature to bombard gaseous hydrogen is 12.5eV.
In the ground state , energy of gaseous hydrogen at room temperature is -13.6 eV.
When the gaseous hydrogen is bombarded with an electron beam , the energy of the gaseous hydrogen becomes
$$E= -13.6eV + 12.5 eV = -1.1eV$$
The orbital energy is related to orbit level (n) as :
$$E = - \frac{13.6}{n^{2}}eV$$
For n = 3 ,
$$E = - \frac{13.6}{(9)^{2}} = -1.5eV$$
The energy is almost equal to the gaseous hydrogen energy. It is possible to conclude that the electron jumped from level n = 1 to level n=3
During they de-excitation , the electron can jump from n = 3 to n = 1 , forming a line of the lyman series of the hydrogen spectrum.
$$\frac{1}{\lambda} = R [ \frac{1}{(n_{1})^{2}}- \frac{1}{(n_{2})^{2}}]$$
Where R = Rydberg constant
$$R = 1.097 \times 10^{7} m^{-1}$$
The wavelength of the radiation when the electron jumps from n = 3 to n = 1 is given as :
$$\frac{1}{\lambda} = R [ \frac{1}{(1)^{2}}- \frac{1}{(3)^{2}}]$$
$$\frac{1}{\lambda} = 1.097 \times 10^{7} [ \frac{1}{(1)^{2}}- \frac{1}{(3)^{2}}]$$
$$\frac{1}{\lambda} = 1.097 \times 10^{7}(1- \frac{1}{9})$$
$$\frac{1}{\lambda}= 1.097 \times 10^{7} \times \frac{8}{9}$$
$$\lambda = \frac{9}{8 \times 1.097 \times 10^{7}}$$
$$\lambda= 102.55 nm$$
The wavelength of the radiation when the electron jumps from n=2 to n = 1 is given as :
$$\frac{1}{\lambda} = R [ \frac{1}{(1)^{2}}- \frac{1}{(2)^{2}}]$$
$$\frac{1}{\lambda} = 1.097 \times 10^{7}(1- \frac{1}{4})$$
$$\frac{1}{\lambda}= 1.097 \times 10^{7} \times \frac{3}{4}$$
$$\lambda = \frac{4}{3 \times 1.097 \times 10^{7}}$$
$$\lambda= 121.54 nm$$
The wavelength of the radiation when the transition takes place from n = 3 to n=2 is given as :
$$\frac{1}{\lambda} = R [ \frac{1}{(2)^{2}}- \frac{1}{(3)^{2}}]$$
$$\frac{1}{\lambda} = R [ \frac{1}{4}- \frac{1}{9}]$$
$$\frac{1}{\lambda}= 1.097 \times 10^{7} \times \frac{9-4}{36}$$
$$\frac{1}{\lambda}= 1.097 \times 10^{7} \times \frac{5}{36}$$
$$\lambda = \frac{36}{5 \times 1.097 \times 10^{7}}$$
$$\lambda= 656.33 nm$$
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence , in the Balmer series , one wavelength i.e 656.33 nm is emitted and in Lyman series , two wavelengths i.e 102.55 and 121.54 nm are emitted.
12.9 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 $1.5 \times 10^{11}m$ with orbital speed $3 \times 10^{4}$m/s. (Mass of earth= $6.0 \times 10^{24}kg$.)
Solution:
Here ,
$$m= 6.0 \times 10^{24}kg$$
$$r = 1.5 \times 10^{11}m$$
$$v= 3 \times 10^{4} m/s$$
Angular momentum of earth
$$L = mvr$$
$$L = 6.0 \times 10^{24} \times 3 \times 10^{4} \times 1.5 \times 10^{11}$$
$$L = 2.7 \times 10^{40} kg m/s$$
According to Bohr's model , angular momentum of earth around the sun is
$$L = n \times \frac{h}{2 \pi}$$
$$mvr = n \times \frac{h}{2 \pi}$$
$$n = \frac{2 \pi mvr}{h}$$
$$n = \frac{2 \pi}{h} \times 2.7 \times 10^{40}$$
$$n = \frac{2 \pi}{6.62 \times 10^{-34}} \times 2.7 \times 10^{40}$$
$$n = 2.57 \times 10^{74}$$