Atoms and Nuclei

Radioactive decay is a random and spontaneous process it depends on unbalancing of nucleus.

$N = {N_0}{e^{ - \lambda t}}$ ..... (B)

$\ln N = - \lambda t + \ln {N_0}$

So, slope $ = - \lambda $ ..... (C)

${t_{1/2}} = {{\ln 2} \over \lambda }$

So ${t_{1/2}} \times \lambda = \ln 2 = $ Constant

K_p > 0

If Q is released $\Rightarrow$ Q > 0

$\Rightarrow$ K_p + Q > 0

Even the particle has to be given kinetic energy greater than magnitude of Q to maintain momentum conservation.

$\Rightarrow$ K + Q > 0

Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.

Also, E₂ $-$ E₁ is the energy of photon

$\Rightarrow$ E₂ $-$ E₁ = hf

$ \Rightarrow f = {{{E_2} - {E_1}} \over h}$

$ - 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}$

$ \Rightarrow {{13.6} \over {{n^2}}} = 3.4$

$ \Rightarrow n = 2$

$ \Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}$

$ = {h \over {2\lambda }}$

$ \Rightarrow \Delta L \simeq 1.05 \times {10^{ - 34}}$ Js

${{dA} \over {dt}} - ( - {\lambda _1}A) + ( - {\lambda _2}A)$

$ \Rightarrow {{dA} \over {dt}} = - ({\lambda _1} + {\lambda _2})A$

$ \Rightarrow {\lambda _{eff}} = {\lambda _1} + {\lambda _2}$

$ \Rightarrow {{\ln 2} \over {{{({t_{1/2}})}_{eff}}}} = {{\ln 2} \over {{{({t_{1/2}})}_1}}} + {{\ln 2} \over {{{({t_{1/2}})}_2}}}$

$ \Rightarrow {({t_{1/2}})_{eff}} = {{4.5 \times 3} \over {7.5}}$ hours = 1.8 hours

${}_{92}{U^{238}}\buildrel {} \over \longrightarrow {}_{82}P{b^{206}} + x\left( {{}_2H{e^4}} \right) + {}_y\left( {{}_{ - 1}{\beta ^0}} \right)$

$238 = 206 + 4x + 0$

$ \Rightarrow 4x = 32 \Rightarrow x = 8$

also, $92 = 82 + 2x - y$

$y = 82 + 16 - 92 = 6$

We know that

$R = {R_0}{A^{1/3}}$

$ \Rightarrow \underbrace {\ln \left( {{R \over {{R_0}}}} \right)}_y=\underbrace {{1 \over 3}}_m\underbrace {\ln (A)}_x$

$\Rightarrow$ Straight line

We know that $v \propto {Z \over n}$

$\Rightarrow$ Required ratio $ = {{{2 \over 3}} \over {{1 \over 3}}}$

$ = 2:1$

$T.E. = {{ - {Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}$

$P.E. = {{ - {Z^2}m{e^4}} \over {4{{\left( {nh{\varepsilon _0}} \right)}^2}}}$

$K.E. = {{{Z^2}m{e^4}} \over {8{{\left( {nh{\varepsilon _0}} \right)}^2}}}$

As electron makes transition to higher level, total energy and potential energy increases (due to negative sign) while the kinetic energy reduces.

An atom based on classical theory of Rutherford's model should collapse as the electrons in continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy. These electrons losing energy should soon fall into heavy nucleus collapsing the whole atom.

²²⁰A $\to$ ¹⁰⁵B + ¹¹⁵C

$\Rightarrow$ Q = [105 $\times$ 6.4 + 115 $\times$ 6.4] $-$ [220 $\times$ 5.6] MeV

$\Rightarrow$ Q = 176 MeV

Total binding energy (without considering repulsions),

${E_b} = [Z{m_p} + (A - Z){m_n} - {m_x}]{c^2}$

Where, $_Z^AX$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $ = {}^z{C_2}$

$\Rightarrow$ This repulsion energy $ \propto \,{{Z(Z - 1)} \over 2} \times {1 \over {4\pi { \in _0}}}{{{e^2}} \over R}$

Where R is the radius of the nucleus

$ \Rightarrow E_b^p - E_b^n \propto Z(Z - 1)$ $\therefore$ there will be no repulsion term for neutrons.

Also, since $R = {R_0}{A^{1/3}}$

$ \Rightarrow E_b^p - E_b^n \propto {A^{ - 1/3}}$

Because of repulsion among protons,

$E_b^p < E_b^n$

Since in $\beta^+$ decay, number of protons decrease $\Rightarrow$ repulsion would decrease

$ \Rightarrow E_b^p$ increases

Order of magnitude of fundamental forces is Gravity ( $10^{-38}$ ); weak nuclear Forces $\left(10^{-13}\right)$; electromagnetic $\left(10^{-2}\right)$; strong nuclear force $\left(10^0\right)$.

∴ Strong nuclear forces are with largest magnitude.

Wavelength of spectral lines of $n$-atom is given as

$ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For shortest wavelength of H - atom spectrum in Balmer series,

$ \begin{aligned} n_1=2 \text { and } & n_2=\infty \\ \therefore \quad \frac{1}{\lambda} & =R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ \frac{1}{\lambda} & =\frac{R}{4} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad \lambda & =\frac{4}{R} \\ & =\frac{4}{1.097 \times 10^7}=3.646 \times 10^{-7} \mathrm{~m} \\ & =3646 \times 10^{-10} \mathrm{~m}=3646 \mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

Total energy released = Total number of fission × Energy released per fission

$ \begin{aligned} & =m \times \frac{N}{M} \times 200 \times 10^6 \times 1.6 \times 10^{-19} \\ & =1 \times 10^{-3} \times \frac{6 \times 10^{23}}{240} \\ & \quad \times 200 \times 10^6 \times 1.6 \times 10^{-19} \\ & =8 \times 10^7 \mathrm{~J} \end{aligned} $

Range of fundamental forces of nature (in ascending order) is

Weak nuclear force (less than $10^{-18} \mathrm{~m}$ );

Strong nuclear force (less than $10^{-15} \mathrm{~m}$ );

Electromagnetic force (very large) ;

Gravity (infinity)

Strength of forces (in ascending order) is

Gravity ( $10^{-38}$ ) :

weak nuclear ( $10^{-13}$ );

electromagnetic $\left(10^{-2}\right)$;

strong nuclear force $\left(10^{\circ}\right)$.

The energy of an electron in $n$th excited state of H -atom is given as

$ E_n=\frac{-13.6}{n^2} \mathrm{eV} $

For fourth excited state,

$ n=5 $

$ \begin{aligned} \therefore \quad E_5 & =\frac{-13.6}{5^2} \\ & =\frac{-13.6}{25} \\ & =-0.54 \mathrm{eV} \end{aligned} $

Given, for Al nucleus

Mass number, $A=27$

$ \begin{aligned} R_0 & =1 \times 10^{-15} \mathrm{~m} \\ \pi & =3 \end{aligned} $

and

We know that, radius of nucleus is given as

$ \begin{aligned} R & =R_0 A^{1 / 3} \\ & =1 \times 10^{-15} \times(27)^{1 / 3} \\ & =3 \times 10^{-15} \mathrm{~m} \end{aligned} $

Volume of aluminium (Al) nucleus,

$ \begin{aligned} V & =\frac{4}{3} \pi R^3 \\ & =\frac{4}{3} \times \pi \times\left(3 \times 10^{-15}\right)^3 \\ & =\frac{4}{3} \times 3 \times 27 \times 10^{-45} \\ & =108 \times 10^{-45} \mathrm{~m}^3 \\ & =1.08 \times 10^{-13} \times\left(10^{-10}\right)^3 \mathrm{~m}^3 \\ & =1.08 \times 10^{-13} \times(1 \mathop {\rm{A}}\limits^{\rm{o}})^3 \\ & \simeq 1 \times 10^{-13} {\mathop {\rm{A^3}}\limits^{\rm{o}} } \end{aligned} $

The nuclear force is a powerful attractive force between nucleons at a distance of about 1 femtometer or $1 \times 10^{-15} \mathrm{~m}$.

∴ Hence, range of nuclear force

$ =10^{-15} \mathrm{~m} $

According to Bohr's model of hydrogen atom, velocity of electron in $n$th orbit is given as

$ v \propto \frac{1}{n} $

For fourth orbit, $n_4=4$

For 9 th orbit, $n_9=9$

$ \begin{array}{rlrl} \therefore & \frac{v_4}{v_9} & =\frac{n_9}{n_4}=\frac{9}{4} \\ \therefore & v_4 & : v_9=9: 4 \end{array} $

Radius of ${ }^{189} \mathrm{O}$,

$ R=R_0(189)^{1 / 3} \quad\left[\because R=R_0 A^{1 / 3}\right] $

∴ Radius of given nucleus,

$ \begin{aligned} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, R^{\prime} =\frac{1}{3} R \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, R^{\prime} =\frac{1}{3} R_0(189)^{1 / 3} \\ &But \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,R^{\prime} =R_0\left(A^{\prime}\right)^{1 / 3} \\ & \Rightarrow \frac{1}{3} R_0(189)^{1 / 3} = R_0\left(A^{\prime}\right)^{1 / 3} \\ & \Rightarrow \quad \frac{1}{3}(189)^{1 / 3} = \left(A^{\prime}\right)^{1 / 3} \\ & \Rightarrow \quad\left(A^{\prime}\right)^{1 / 3} = \frac{1}{3}(189)^{1 / 3} \end{aligned} $

Cubing both side, we get

$ A^{\prime}=\frac{1}{3^3} \cdot 189=\frac{189}{27}=7 $

We know that, wavelength of spectral line in Balmer series is given as,

$ \frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

where $n_1=2$ and $n_2=3,4 \,\,\,\ldots.(i)$

$ \Rightarrow \quad \frac{1}{\lambda}=R_H\left[\frac{1}{2^2}-\frac{1}{n_2^2}\right] $

For maximum wavelength $n_2=3$

Form Eq. (i), we get

$ \begin{aligned} \frac{1}{\lambda_{\max }} & =R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ & =\frac{5}{36} R_H \\ \Rightarrow \quad \lambda_{\max } & =\frac{36}{5 R_H} \\ & =\frac{36}{5 \times 1 \times 10^7}=7.2 \times 10^{-7} \mathrm{~m} \\ \Rightarrow \quad \lambda_{\max } & =7200 \times 10^{-10} \mathrm{~m} \\ & =7200\mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

For minimum wavelength,

$ n_2=\infty $

$ \begin{aligned} \frac{1}{\lambda_{\min }} & =R_H\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{R_H}{4} \\ \Rightarrow \quad \lambda_{\min } & =\frac{4}{R_H}=\frac{4}{1 \times 10^7} \end{aligned} $

$ \begin{aligned} & =4 \times 10^{-7} \mathrm{~m} \\ \Rightarrow \quad \lambda_{\min } & =4000 \times 10^{-10} \mathrm{~m}=4000 \mathop {\rm{A}}\limits^{\rm{o}} \\ \therefore \lambda_{\max }-\lambda_{\min } & =7200-4000=3200\mathop {\rm{A}}\limits^{\rm{o}} \end{aligned} $

We know that, radius of nucleus in terms of mass number $(A)$ is given as

$ \begin{aligned} R & =R_0\left(A^{1 / 3}\right) \\ \Rightarrow \quad \frac{R_2}{R_1} & =\left(\frac{A_2}{A_1}\right)^{\frac{1}{3}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)\end{aligned} $

But, $A_2=A_1+2 \%$ of $A_1$

$ =A_1+\frac{2}{100} A_1=\frac{102}{100} A_1=1.02 A_1 $

From Eq (i), we get

$ \begin{aligned} \frac{R_2}{R_1} & =\left(\frac{1.02 A_1}{A_1}\right)^{\frac{1}{3}}=(1.02)^{1 / 3} \\ R_2 & =R_1(1.02)^{1 / 3} \\ R_2 & =1.0067 R_1=R_1+0.0067 R_1 \\ & =R_1+0.67 \% \text { of } R_1 \\ & =R_1+\frac{2}{3} \% \text { of } R_1 \end{aligned} $

Hence, $R_2$ is larger than $R_1$ by $\frac{2}{3} \%$.

Weak nuclear force is responsible for radioactive decays and hence these forces are also responsible for beta decay.

We know that, frequency of spectral series of hydrogen atom is given as

$ v=R C \quad\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] $

For series limit of Balmer series,

$ \begin{aligned} & & n_1 & =2 \text { and } n_2=\infty \\ \therefore & & v_B & =R C\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ \Rightarrow & & v_B & =\frac{R C}{4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

For the series limit of Brackett series,

$ n_1=4 \text { and } n_2=\infty $

$ \therefore \quad v_B^{\prime}=R C\left(\frac{1}{4^2}-\frac{1}{\infty^2}\right)=\frac{R C}{16} $

Given, I amu $=1.6 \times 10^{-27} \mathrm{~kg}$

Since, nuclear density is independent of mass number, the nuclear density of nucleus ${ }_{30}^{60} \mathrm{X}$ is given as :

$ \rho=\frac{3 m}{4 \pi R_0^3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $m$ is average mass of nucleon.

$ \begin{aligned} & \text { Average mass of nucleon }=\frac{m_n+m_p}{2} \\ & \qquad \begin{aligned} & =\frac{1.0073+1.0086}{2} \mathrm{amu} \\ & =\frac{2.0159}{2}=1.00795 \mathrm{amu} \\ & =1.00795 \times 1.6 \times 10^{-27} \mathrm{~kg} \\ \Rightarrow \quad m & =1.61272 \times 10^{-27} \mathrm{~kg} \end{aligned} \end{aligned} $

∴ Putting the values in Eq. (i), we get

$ \begin{aligned} \rho & =\frac{3 \times 1.61272 \times 10^{-27}}{4 \times 314 \times\left(1.2 \times 10^{-15}\right)^3} \\ & =2.2 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3} \end{aligned} $

Let $A$ be the mass number and $R$ be the radius of a nucleus. If $m$ be the average mass of a nucleon, then mass of nucleus, $M=m A$

Volume of nucleus, $V=\frac{4}{3} \pi R^3$

∴ Nuclear density,

$ \begin{aligned} \rho & =\frac{\text { Mass of nucleus }(M)}{\text { Volume }(V)} \\ & =\frac{m A}{\frac{4}{3} \pi R^3}=\frac{3 m A}{4 \pi R^3} \\ & =\frac{3 m A}{4 \pi\left(R_0 A^{\frac{1}{3}}\right)^3} \quad\left(\because R=R_0 A^{1 / 3}\right) \\ & =\frac{3 m A}{4 \pi R_0^3 A}=\frac{3 m}{4 \pi R_0^3} \end{aligned} $

Hence, nuclear density does not depend on mass number.

Therefore, with increase of mass number A, nuclear density does not change.

Change in angular momentum of the electron in H-atom when it is excited from 2nd orbit to 4th orbit is given as

$\Delta L=L_2-L_I$

Where, $L_2$ is angular momentum of 4th excited state $(n=5)$ and $L_1$ is angular moment of 2nd excited state $(n=3)$.

$\begin{aligned} \therefore \Delta L & =\frac{n_2 h}{2 \pi}-\frac{n_1 h}{2 \pi} \\ & =\frac{h}{2 \pi}\left[n_2-n_1\right]=\frac{6.64 \times 10^{-34}}{2 \times 314}[5-4] \\ & =2.11 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned}$

We know that, the radius $(R)$ of nucleus is given as

$R=R_0 A^{1 / 3} \text { or } R \propto A^{1 / 3}$

Where,

$\begin{aligned} R_0 & =\text { constant } \\ A & =\text { mass number } \end{aligned}$

$\therefore$ Nuclear density, $\rho=\frac{\text { mass }}{\text { volume }}=\frac{m A}{\frac{4}{3} \pi R^3}$

(here, $m$ is mass of each nucleon)

$\begin{aligned} \Rightarrow \quad \rho & =\frac{m A}{\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3}=\frac{m A}{\frac{4}{3} \pi R_0^3 A} \\ \rho & =\frac{3 m}{4 \pi R_0^3} \end{aligned}$

Hence, it is clear from above formula, nuclear density is independent of mass number $A$.

The binding energy, $E=\Delta m c^2$

i. e $E \propto \Delta m$

Where, $\Delta m=$ mass defect

Energy is released when heavy nuclei undergo transmutation into light nuclei.

For a carbon sample $(\mathrm{C}^{14})$,

$T_{1 / 2}=5730 \text { year }$

Decay constant,

$\begin{aligned} & K=\frac{0.693}{T_{1 / 2}}\\ & =\frac{0.693}{5730} \\ & =1.209 \times 10^{-4} / \text { year } \end{aligned}$

The rate of counts is proportional to the number of $\mathrm{C}^{14}$ atom in the sample

$N_0=100, N=75$

The age of the sample is given as,

$\begin{aligned} t & =\frac{2.303}{K} \log \frac{N_0}{N}=\frac{2.303}{K} \log \frac{1}{0.75} \\ & =\frac{2303}{1.209 \times 10^{-4}} \log \frac{100}{75} \\ & =2456 \times 10^3 \text { years } \\ & =2456 \text { years } \end{aligned}$

Energy of hydrogen atom in ground state,

$E_0=-13.6 \mathrm{~eV}$

Energy of electron in excited state $(n=4)$,

$\begin{aligned} E^{\prime}=\frac{-13.6}{n^2} & \\ & =\frac{-13.6}{4^2}=\frac{-13.6}{16}=-0.85 \mathrm{eV} \end{aligned}$

Potential energy of the electron in the excited state $(n=4)$,

$\begin{aligned} U & =2 E^{\prime} \\ & =2 \times(-0.85 \mathrm{eV}) \\ & =-1 \cdot 7 \mathrm{eV} \end{aligned}$

We know that, radius (R) of atomic nucleus and mass number (A) are related as

$\begin{array}{ll} & R=R_0 A^{1 / 3} \\ \Rightarrow & R \propto A^{1 / 3} \\ \Rightarrow & \frac{R_2}{R_1}=\left(\frac{A_2}{A_1}\right)^{1 / 3} \\ \Rightarrow & \frac{A_2}{A_1}=\left(\frac{R_2}{R_1}\right)^3 \end{array}$

Given, $A_1=64, R_1=4.8$ fermi, $R_2=6$ fermi

Putting these values in above equation, we get

$\begin{aligned} & \frac{A_2}{64}=\left(\frac{6}{4.8}\right)^3=\left(\frac{5}{4}\right)^3 \\ & \Rightarrow \quad \frac{A_2}{64}=\frac{125}{64} \\ & \Rightarrow \quad A_2^{\prime}=125 \\ & \end{aligned}$

Energy of electron in $n$th orbit of hydrogen atom,

$E=\frac{-13.6}{n^2} \mathrm{~eV}$

energy of electron in ground state ($n=1$),

$E_0=\frac{-13.6}{1^2}=-13.6 \mathrm{~eV}$

for second excited state, $n=3$

$\therefore$ energy of electron in H -atom in second excited state

$E^{\prime}=\frac{-13.6}{3^2}=-1.51 \mathrm{~eV}$

Energy required to excite the electron in hydrogen atom to second excited state

$=E^{\prime}-E_0$

$\begin{aligned} & =-1.51-(-13.6)=1209 \mathrm{eV} \\ & \simeq 12 \mathrm{~eV} \end{aligned}$

Energy of ionised electron of $(n=\infty) \mathrm{H}$ - atom.

$E^{\prime \prime}=\frac{-13.6}{\infty}=0 \mathrm{~eV}$

energy required to ionise the electron

$=E^{\prime \prime}-E_0=0-(-13.6)=13.6 \mathrm{~eV}$

Radius of nucleus (R) having mass number (A) is given as,

$R=R_0 A^{1 / 3}$

where, $R_0$ is constant.

Taking $\log$ both side, we have

$\begin{aligned} & \ln R=\ln R_0 A^{1 / 3} \\ & \Rightarrow \ln R=\ln R_0+\ln A^{1 / 3} \\ & \Rightarrow \ln R=\ln R_0+\frac{1}{3} \ln A \\ & \Rightarrow \ln R=\frac{1}{3} \ln A+\ln R_0 \end{aligned}$

this is in the form of equation of straight line

$y=m x+c$

where, $y=\ln R$

$m=\frac{1}{3}, x=\ln A$

and $\quad c=\ln R_0$

From radioactive decay law,

$ - {{dN} \over {dt}} \propto N$ or $ - {{dN} \over {dt}} = \lambda N$

Thus, $A = - {{dN} \over {dt}}$ or $A = \lambda N$

$ \Rightarrow A = \lambda {N_0}{e^{ - \lambda t}}$ ...... (i)

Where, ${A_0} = \lambda {N_0}$ is the activity of the radioactive material at time, t = 0

At time, ${A_1} = {A_0}\,{e^{ - \lambda {t_1}}}$ ..... (ii)

At time, ${A_2} = {A_0}\,{e^{ - \lambda {t_2}}}$ ..... (iii)

On dividing Eq. (ii) by Eq. (iii), we have

${{{A_1}} \over {{A_2}}} = {{{e^{ - \lambda {t_1}}}} \over {{e^{ - \lambda {t_2}}}}} = {e^{ - \lambda ({t_1} - {t_2})}}$

$ \Rightarrow {A_1} = {A_2}\,{e^{ - \lambda ({t_1} - {t_2})}}$