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 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.)

Solution :

(a) The sizes of the atoms taken in Thomson's model and Rutherford's model have the $\underline{same \ order \ of \ magnitude.}$$\\$ (b) In the ground state of $\underline{Thomson's \ model}$, the electrons are in stable equilibrium. However, in Rutherford's model, the electrons always experience a net force.$\\$ (c) A classical atom based on $\underline{Rutherford's \ model}$ is doomed to collapse.$\\$ (d) An atom has a nearly continuous mass distribution in $\underline{Thomson's \ model}$, but has a highly non-uniform mass distribution in ${Rutherford's \ model.}$$\\$ (e) The positively charged part of the atom possesses most of the mass in $\underline{both the models.}$

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 :

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 is less than the mass of incident $\alpha $ - particles Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the $\alpha$- particles would not bounce back if solid hydrogen is used in the $\alpha -$ particle scattering experiment and so we cannot determine size of the hydrogen

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

Solution :

Rydberg’s Formula is given as:$\\$ $\dfrac{hc}{\lambda} =21.76*10^{-19}\left[\dfrac{1}{n_1^2} -\dfrac{1}{n_2^2}\right]$$\\$ Where, $h =$ Planck’s constant $= 6.6*10^{-34} Js$$\\$ $c=$ Speed of light =$3*10^8 m/s$$\\$ ($ n _1$ and $n _2$ are integers)$\\$ The shortest wavelength present in the Paschen series of the spectral lines is for values $n_1 = 3$ and $n_2 =\infty$$\\$ $\dfrac{hc}{\lambda}=21.76*10^{-19}\left[\dfrac{1}{(3)^2}-\dfrac{1}{(\infty)^2}\right]$$\\$ $\lambda =\dfrac{6.6*10^{-34}*3*10^8*9}{21.76*10^{-19}} \\ =8.189*10^{-7}m\\ 818.9 nm$

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?

Solution :

Separation of two energy levels in an atom,$\\$ $E=2.3eV\\ =2.3*1.6*10^{-19}\\ =3.68*10^{-19}J$$\\$ Let $\theta $ be the frequency of radiation emitted when the atom transits from the upper level to the lower level. $\\$ We have the relation for energy as:$\\$ $E=hv$$\\$ Where,$\\$ $h=$ Planck's constant$=6.62*10^{-34} Js$$\\$ $\therefore v=\dfrac{E}{h}\\ =\dfrac{3.68*10^{-19}}{6.62*10^{-32}}\\ =5.55*10^{14} Hz$$\\$ Hence, the frequency of the radiation is $5.6 * 10^{14} Hz .$

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?

Solution :

Ground state energy of hydrogen atom, $E =- 13.6 eV$$\\$ This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy $=-E =-(-13.6 )=13.6 e V$$\\$ Potential energy is equal to the negative of two times of kinetic energy. Potential energy = $-2 *( 13.6 )=- 27.2 e V$

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.

Solution :

For ground level, $n _1 = 1$$\\$ Let $E_ 1$ be the energy of this level. It is known that $E _1$ is related with $n_ 1$ as:$\\$ $E_1=\dfrac{-13.6}{n_1^2}eV\\ =\dfrac{-13.6}{1^2}=-13.6 eV$$\\$ The atom is excited to a higher level, $n_ 2 = 4$$\\$ Let $E _2$ be the energy of this level.$\\$ $\therefore E_2=\dfrac{-13.6}{n_2^2}eV\\ =\dfrac{-13.6}{4^2}=-\dfrac{13.6}{16}eV$$\\$ The amount of energy absorbed by the photon is given as:$\\$ $E = E_ 2 - E_ 1\\ =\dfrac{-13.6}{16}-(-\dfrac{13.6}{1})\\ =\dfrac{13.6*15}{16}eV\\ =\dfrac{13.6*15}{16}*1.6*10^{19}=2.04*10^{-18}J$$\\$ For a photon of wavelength $\lambda $ , the expression of energy is written as:$\\$ $E=\dfrac{hc}{\lambda}$$\\$ Where,$\\$ h=Planck's constant $=6.6*10^{-34}Js$$\\$ c=Speed of light =$ 3*10^{8}m/s$$\\$ $\therefore \lambda =\dfrac{hc}{E}\\ =\dfrac{6.6*10^{-34}*3*10^8}{2.04*10^{-18}}\\ =9.7*10^{-8}m=97 nm$$\\$ And, frequency of a photon is given by the relation,$\\$ $v=\dfrac{c}{\lambda}\\ =\dfrac{3*10^8}{9.7*10^{-8}} \approx 3.1*10^{15}Hz$$\\$ Hence, the wavelength of the photon is $ 97 nm $ while the frequency is $ 3.1 * 10^{15} Hz .$$\\$

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.$\\$

Solution :

(a) Let $v _1$ be the orbital sped of the electron in a hydrogen atom in the ground state level $n_ 1 = 1$ . For charge $( e )$ of an electron, $v _1$ is given by the relation,$\\$ $v_1=\dfrac{e^2}{n_14\pi \in_0(h/2\pi)}\\ =\dfrac{e^2}{2\in_0h}$$\\$ Where ,$\\$ $e=1.6*10^{-19}C$$\\$ $\in_0=$ Permittivity of free space $=8.85*10^{-12}N^{-1}C^2m^{-2}$$\\$ h= Planck's constant $ =6.62*10^{-34}Js\\ \therefore v_1=\dfrac{(1.6*10^{-19})^2}{2*8.85*10^{-12}*6.62*10^{-34}}\\ =0.0218*10^8=2.18*10^6 m/s$$\\$ For level $n_ 2 = 2$ , we can write the relation for the corresponding orbital speed as:$\\$ $v_2=\dfrac{e^2}{n_22\in_0h}\\ =\dfrac{(1.6*10^{-19})^2}{2*2*8.85*10^{-12}*6.62*10^{-34}}\\ =1.09*10^6 m/s$$\\$ And, for $n_ 3 = 3$ , we can write the relation for the corresponding orbital speed as:$\\$ $v_3=\dfrac{e^2}{n_32\in_0h}\\ =\dfrac{(1.6*10^{-19})^2}{3*2*8.85*10^{-12}*6.62*10^{-34}}\\ =7.27*10^5 m/s$$\\$ 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:$\\$ $T_1=\dfrac{2 \pi r_1}{v_1}$$\\$ Where,$\\$ $r_1=$ Radius of the orbit$\\$ $=\dfrac{n_1^2 h^2 \in_0}{\pi m e ^2}$ $\\$

$h=$ Planck's contant = $ 6.62*10^{-34}Js$$\\$ $e=$ Charge on a electron=$ 8.85*10^{-12}N^{-1}C^2 m^{-2}$$\\$ $m=$ Mass of an electron $=9.1*10^{-31}kg$$\\$ $\therefore T_1=\dfrac{2 \pi r_1}{v_1}\\ =\dfrac{2\pi*(1)^2*(6.62*10^{-34})^2*8.85*10^{-12}}{2.18*10^6*\pi * 9.1*10^{-31}*(1.6*10^{-19})^2}\\ =15.27*10^{-17}=1.527*10^{-16}s$$\\$ For level $n_ 2 = 2$, we can write the period as:$\\$ $T_2=\dfrac{2 \pi r_2}{v_2}$$\\$ Where,$\\$ $r_2=$ Radius of the electron in $ n_2=2$$\\$ $=\dfrac{(n_2)^2h^2 \in_0}{\pi me^2}\\ \therefore T_2=\dfrac{2\pi r_2}{v_2}\\ =\dfrac{2\pi * (2)^2*(6.62*10^{-34})^2*8.85*10^{-12}}{1.09 * 10^6* \pi * 9.1 * 10^{31}*(1.6*10^{-19})^2}\\ =1.22*10^{-15}s$$\\$ And, for level $n_ 3 = 3$, we can write the period as:$\\$ $T_3=\dfrac{2\pi r_3}{v_3}$$\\$ Where,$\\$ $r_3=$ Radius of the electron in $ n_3=3$$\\$ $=\dfrac{(n_3)^2h^2 \in_0}{\pi me^2}\\ \therefore T_3=\dfrac{2 \pi r_3}{v_3}\\ =\dfrac{2\pi*(3)^2*(6.62*10^{-34})^2*8.85*10^{-12}}{7.27* 10^5*\pi*9.1*10^{-31}*(1.6*10^{-19})^2}\\ =4.12 * 10^{-15}s$$\\$ 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.

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?

Solution :

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: $r _2 =( n )^2 r_ 1\\ = 4 * 5.3 * 10 ^{- 11} =2.12 * 10 ^{-10} m$$\\$ For $n = 3$, we can write the corresponding electron radius as: $r_ 3 =( n)^2 r_ 1 \\ = 9 * 5.3 * 10 ^-{ 11 }= 4.77 * 10 ^{- 10} m$$\\$ 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.

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

Solution :

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.$$\\$ When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes $- 13.6 + 12.5 eV _i e ,-1.1eV.$$\\$ Orbital energy is related to orbit level (n) as:$\\$ $E=\dfrac{-13.6}{(n)^2}eV$$\\$ For $ n=3,E=\dfrac{-13.6}{9}=-1.5eV$$\\$ 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:$\\$ $\dfrac{1}{\lambda }=R_y(\dfrac{1}{1^2}-\dfrac{1}{n^2})$$\\$ Where,$\\$

$R_y=$ Rydberg constant $= 1.097 * 10 ^7 m ^{- 1}$$\\$ $\lambda =$ Wavelength of radiation emitted by the transition of the electron For $n = 3$ , we can obtain $\lambda $ as:$\\$ $\dfrac{1}{\lambda}=1.097*10^7(\dfrac{1}{1^2}-\dfrac{1}{3^2})\\ =1.097*10^7(1-\dfrac{1}{9})=1.097*10^{7}*\dfrac{8}{9}\\ \lambda =\dfrac{9}{8*1.097*10^7}=102.55nm$$\\$ If the transition takes place from $n = 3$ to $n = 2$ , and then from $n =2$ to$ n = 1$, then the wavelength of the radiation 3emitted in transition from $n = 3$ to $n = 2$ is given as :$\\$ $\dfrac{1}{\lambda}=1.097*10^7(\dfrac{1}{2^2}-\dfrac{1}{3^2})\\ =1.097*10^7(\dfrac{1}{4}-\dfrac{1}{9})=1.097*10^7*\dfrac{5}{36}\\ \lambda =\dfrac{36}{5*1.097*10^7}=656.33 nm$$\\$ This radiation corresponds to the Balmer series of the hydrogen spectrum.$\\$ The wavelength of the radiation when transition takes place from $n = 2$ to $n =1$ ,is given as:$\\$ $\dfrac{1}{\lambda}=1.097*10^7(\dfrac{1}{1^2}-\dfrac{1}{2^2})\\ =1.097*10^7(1-\dfrac{1}{4})=1.097*10^7*\dfrac{3}{4}\\ \lambda =\dfrac{4}{1.097*10^7*3}=121.54 nm$$\\$ Hence, in Lyman series, two wavelengths i.e.,$102.5 nm $ and $121.5nm $are emitted. And in the Blamer series, one wavelength ie., $656.33 nm $ is emitted.

10   In accordance with the Bohr’s model, find the quantum number that characterizes 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 $=6.0 *10^{ 24} kg . )$

Solution :

Radius of the orbit of the Earth around the Sun,$ r = 1.5 * 10^{ 11} m$$\\$ Orbital speed of the Earth, $v =3 * 10^ 4 m / s$$\\$ Mass of the Earth, $m = 6.0 * 10^{ 24} kg$$\\$ According to Bohr’s model, angular momentum is quantized and given as:$\\$ $mvr=\dfrac{nh}{2\pi}$$\\$ Where, $\\$ $h=$ Planck's Constant=$6.62*10^{-34}Js$$\\$ $n=$ Quantum Number$\\$ $\therefore n=\dfrac{mvr^2\pi}{h}\\ =\dfrac{2\pi*6*10^{24}*3*10^4*1.5*10^{11}}{6.62*10^{34}}\\ =25.61*10^{73}=2.6*10^{74}$$\\$ Hence, the quantum number that characterizes the Earth’s Revolution is $2.6 * 10^{ 74}$