Atoms and Nuclei

As we know that,

By using de-Broglie wavelength,

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}$

where, $\lambda$ is wavelength of $X$-ray, $h$ is Planck's constant i.e. $6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}, p$ is, momentum, $m$ is mass, $E$ is energy, $e$ is charge and $V$ is electric potential.

$\therefore \quad \lambda \propto \frac{1}{\sqrt{V}} $

By using Rydberg's formula

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

For Lyman series,

$\begin{array}{rlrl} n_1 & =1 \text { and } n_2=2 \\ \therefore \quad \quad \frac{1}{\lambda} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R\left(\frac{1}{1}-\frac{1}{4}\right)=R\left(\frac{3}{4}\right) \\ \Rightarrow \quad \lambda & =4 / 3 R=1215 \times 10^{-10} \mathrm{~m} \\ & =\frac{4}{3 \times 1.097 \times 10^7}=1215 \mathop A\limits^o\end{array}$

As we know that,

In case of $(-\beta)$ - decay we get $+e_1^0$ positron, so atomic number of reactant increases get by 1

and in case of $\gamma$-decay we get only gamma-ray and no change in atomic number as well as mass.

$\therefore \mathrm{Np}_{93}^{239} \longrightarrow \mathrm{Pu}_{94}^{239}+\beta^{-}+\gamma^{-}$ is possible whereas (a), (b) and (d) are not possible.

As we know that, angular momentum of electron, $L=nh/2\pi$.

$\therefore$ L is integral multiple of $h/2\pi$.

As we know that, nuclear density of any atom $=2\times10^{17}$ kg/m$^3$

$\therefore$ Option (d) is correct.

Ionisation potential of hydrogen $=13.6 \mathrm{~V}$

Let energy $E$ is required to excite electron to 1st excited state $n=2$

Since, $E=\frac{13.6}{n^2} Z^2 \mathrm{~eV}$

$\Rightarrow \quad E=\frac{13.6}{2^2}=\frac{13.6}{4}=3.4 \mathrm{~eV}$

As we know that in case of fussion reaction energy released is proportional to mass.

$\therefore$ When separated nucleons combine some mass gets converted to energy.

Hence, mass of stable nuclei will always be less than separated nucleons.

Given, potential energy between proton and electron,

$U=\frac{K e^2}{3 R^3}$

As we know that,

Force, $F=\frac{-d U}{d R}$

$\therefore \quad F=-\frac{K e^2}{3} \frac{d}{d R} R^{-3} \Rightarrow \frac{m v^2}{R}=-\frac{K e^2}{3} \cdot(-3) R^{-4}$

where, $m$ is mass,

$v$ is speed and $R$ is radius.

$\therefore \quad \frac{m v^2}{R}=\frac{K e^2}{R^4} $

$\Rightarrow \quad R^3=\frac{K e^2}{m v^2}$ ...... (i)

From uncertainty principle,

$\begin{aligned} L & =m v R=\frac{n h}{2 \pi} \\ \Rightarrow \quad v & =\frac{n h}{2 \pi m R} \end{aligned}$

Substituting the value in Eq. (i), we get

$\begin{aligned} & R^3=\frac{K e^2}{m \cdot \frac{n^2 h^2}{4 \pi^2 m^2 R^2}} \\ \therefore \quad & R^3=\frac{K e^2 \cdot 4 \pi^2 m^2 R^2}{m n^2 h^2} \\ \therefore \quad & R=\frac{4 \pi^2 \cdot K e^2 \cdot m}{n^2 h^2} \end{aligned}$