# Atoms

## Physics

### NCERT

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?

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.$\\$

(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 $2.6 * 10^{ 74}$