Atoms and Nuclei

As the ${ }_5^{12} B$ is going under $\beta$-decay to form $_6^{12}C$. We can write the balanced equation as follows

$ { }_5^{12} \mathrm{~B} \rightarrow{ }_6^{12} \mathrm{C}+e^{-1}+\bar{\nu} $

The $Q$-value of this reaction is given by

$ \begin{aligned} Q & =\left[m\left({ }_5^{12} \mathrm{~B}\right)-m\left({ }_6^{12} \mathrm{C}\right)\right] c^2 \\\\ & =[12.041-12.0] \times 931.5=13.041 \mathrm{MeV} \end{aligned} $

The energy $Q=13.041 \mathrm{MeV}$ is released in the reaction. Out of this energy, 4.041 $\mathrm{MeV}$ is used to excite ${ }_6^{12} \mathrm{C}$ to its excited state ${ }_6^{12} \mathrm{C}^*$. Thus, the kinetic energy available to the $\beta$-particle $\left(K_\beta\right)$ and the antineutrino $\left(K_{\bar{\nu}}\right)$ is $K_\beta+K_{\bar{\nu}}=13.041-4.041=9 \mathrm{MeV}$. In $\beta-$ decay, the kinetic energy of the $\bar{\nu}$ can vary from zero to a maximum value. Hence, the maximum kinetic energy of the $\beta$-particle is $K_{\beta, \max }=9 \mathrm{MeV}$ (when $K_{\bar{\nu}}=0$ ).

Law of radioactivity : $N = {N_0}{e^{ - \lambda t}}$ where $\lambda$ = decay constant

Activity $\left| A \right| = \left| {{{ - dN} \over {dt}}} \right| = {N_0}\lambda {e^{ - \lambda t}}$

Rate of activity $R = {{d|A|} \over {dt}} = {N_0}{\lambda ^2}{e^{ - \lambda t}}$

At t = 0, A₁ = A₂. Therefore,

${N_{OP}}{\lambda _P} = {N_{OQ}}{\lambda _Q}$

At $t = 2\tau ,\,{{{R_P}} \over {{R_Q}}} = {\left( {{{{\lambda _P}} \over {{\lambda _Q}}}} \right)^2}\left( {{{{N_{OP}}} \over {{N_{OQ}}}}} \right){{{e^{ - \lambda P(25)}}} \over {{e^{ - \lambda Q(25)}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{({\lambda _Q} - {\lambda _P})25}}$

Since mean life is given by $\tau = {1 \over \lambda }$.

Therefore,

${{{R_P}} \over {{R_Q}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{{{\left( {{1 \over {25}} - {1 \over 5}} \right)}^{25}}}} = {{{\lambda _P}} \over {{\lambda _Q}}}{e^{ - 1}}$

${{{R_P}} \over {{R_Q}}} = {{{\lambda _\{ }} \over {{\lambda _Q}}}{1 \over e} = {n \over e}$

$n = {{{\lambda _P}} \over {{\lambda _Q}}} = {{2\tau } \over \tau } = 2$

Energy of the incident photon $ = hf = {{hc} \over \lambda } = {{1242} \over {90}} = 13.8$ eV. Since after ionisation, electron is ejected with some kinetic energy. By energy conservation, we get

Energy (photon) = Kinetic energy (electron) + $\Delta$E

Transition energy from nth orbit to n $\to$ $\infty$. Therefore,

13.8 = 10.4 + $\Delta$E

$\Rightarrow$ $\Delta$E = 3.4 eV

From Bohr's theory,

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

Half life of radioactive material = T years

Let amount of radioactive material as fuel at the beginning be N₀ and corresponding power produced by it be P₀.

According to question,

Power requirement of the village

= 12.5% of ${P_0} = {{{P_0}} \over 8}$

Since, after each T year, power will be half, i.e.,

${P_0}\buildrel T \over \longrightarrow {{{P_0}} \over 2}\buildrel T \over \longrightarrow {{{P_0}} \over 4}\buildrel T \over \longrightarrow {{{P_0}} \over 8}$

Total time upto which the plant can meet the village's need = 3T years = nT years

$\therefore$ n = 3

Decay constant, $\lambda = {{\ln 2} \over {{T_{1/2}}}} = {{0.693} \over {1386\,s}} = 5 \times {10^{ - 4}}\,s$

According to radioactive decay, $N = {N_0}{e^{ - \lambda t}}$

${N \over {{N_0}}} = {e^{ - 5 \times {{10}^{ - 4}} \times 80}}$ or ${N \over {{N_0}}} = {e^{ - 0.04}}$

Fraction of nuclei decayed $ = {{{N_0} - N} \over {{N_0}}} = 1 - {N \over {{N_0}}}$

$ = 1 - {e^{ - 0.04}} = 1 - 0.96 = 0.04 = 4\% $

The activity of a radioactive substance, having a decay constant $\lambda$ and number of nuclei N at time t, is given by

$A = \left| {dN/dt} \right| = \lambda N = \lambda {N_0}{e^{ - \lambda t}}$ ..... (1)

Take logarithm on both sides of equation (1) to get

$\ln \left| {dN/dt} \right| = \ln (\lambda {N_0}) - \lambda t$ ...... (2)

Thus, the graph between t and $\left| {dN/dt} \right|$ is a straight line with slope $ - \lambda $.

Slope $ = - \lambda = {{3 - 4} \over {6 - 4}}$ (From graph) or $\lambda = {1 \over 2}$ year^$-$1

Half life ${T_{1/2}} = {{0.693} \over \lambda } = 2 \times 0.693$ years = 1.386 years

4.16 years is approximately 3 half-lives

Nuclei will decay by a factor of 2³ = 8

$\therefore$ p = 8

The $n^{\text {th }}$ line in Lyman series corresponds to transition $(n+1) \rightarrow 1$. The wavelength of this transition is given by

$ \begin{aligned} \frac{1}{\lambda} & =\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right| \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \mathrm{R} & =1.097 \times 10^7 \mathrm{~m}^{-1} \text { and } z=11 . \end{aligned} $

The angular momemtum in $n^{\text {th }}$ orbit is given by

$ m v r=\frac{n h}{2 \pi} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

The de-Broglie wavelength of electron in $(n+1)^{\text {th }}$ orbit is

$ \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{2 \pi r}{(n+1)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

$ \begin{aligned} &\text { The radius of }(n+1) \text { th orbit is }\\ &r=a_0(n+1)^2 / z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

where, $a_0=0.5269 \times 10^{-10} m$ is Bohr's radius.

Substitute $r$ from equation (iv) into equation (iii) and then substitute into equation (i) to get.

$ \begin{aligned} \lambda & =\frac{2 \pi\left(a_0(n+1)^2\right) / z}{(n+1)} \\ & =\frac{2 \pi a_0(n+1)}{z} \end{aligned} $

From (i)

$ \begin{aligned} & \frac{\frac{1}{\frac{2 \pi a_0(n+1)}{z}}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right|}{\frac{z}{2 \pi a_0(n+1)}=\mathrm{R} z^2\left|\frac{1}{1^2}-\frac{1}{(n+1)^2}\right|} \\ & \frac{z}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \end{aligned} $

$ \begin{aligned} &=1.09 \times 10^7 \times z^2\left[1-\frac{1}{(n+1)^2}\right]\\ &\text { Put } z=11\\ &\begin{aligned} & \frac{11}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \\ & \quad=1.09 \times 10^7(11)^2\left[1-\frac{1}{(n+1)^2}\right] \\ & \frac{1}{2 \pi\left(0.529 \times 10^{-10}\right)(n+1)} \\ & \quad=1.09 \times 10^7(10)\left[1-\frac{1}{(n+1)^2}\right] \\ & \frac{1}{2 \pi\left(0.529 \times 10^{-10}\right)\left(1.09 \times 10^7\right)(11)} \\ & \quad=\left[1-\frac{1}{(n+1)^2}\right][n+1] \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{1}{2 \pi(0.529)\left(10^{-10}\right)\left(1.09 \times 10^7\right)(11)} \\ & =\frac{(n+1)^2-1}{(n+1)^2} \cdot(n+1) \end{aligned} $

$ \begin{aligned} \frac{1}{2 \pi(0.529)(1.09)(11)\left(10^{-3}\right)} & =\frac{(n+1)^2-1}{(n+1)} \\ \frac{1}{39.852 \times 10^{-3}} & =\frac{(n+1)^2-1}{(n+1)} \\ \frac{1000}{39.852} & =\frac{(n+1)^2-1}{(n+1)} \\ 25.09 & =\frac{(n+1)^2-1}{n+1} \\ \Rightarrow \quad 25 n+25=n^2+2 n & +1-1 \\ & \quad \text { (approximate) } \end{aligned} $

$ \begin{aligned} & \quad \begin{aligned} 25 n & +25=n^2+2 n \\ n^2-23 n & -25=0 \\ n & =\frac{-(-23) \pm \sqrt{(-23)^2-4.1 .(-25)}}{2.1} \\ & =\frac{23 \pm \sqrt{529+100}}{2}=\frac{23 \pm \sqrt{629}}{2} \\ n & =\frac{23+\sqrt{629}}{2}=\frac{23-\sqrt{629}}{2} \\ n & =\frac{23+25}{2}=\frac{23-25}{2} \\ \quad n & =24, n=1 \\\therefore n & =24 \end{aligned} \end{aligned} $