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

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

Kinetic energy of a proton is given by

$E = {1 \over 2}{m_p}v_p^2$

$ \Rightarrow E = {{{p^2}} \over {2{m_p}}}$ [$\because$ $p = mv$]

$\therefore$ Wavelength of proton is, ${\lambda _1} = {h \over p}$

$ \Rightarrow {\lambda _1} = {h \over {\sqrt {2{m_p}E} }}$ ..... (i)

Now, energy of photon is given by

$E = h{v_2} \Rightarrow E = {{hc} \over {{\lambda _2}}}$

$ \Rightarrow {\lambda _2} = {{hc} \over E}$ ....... (ii)

where, $\lambda$₂ is wavelength of photon.

On dividing Eq. (ii) by Eq. (i), we get

${{{\lambda _2}} \over {{\lambda _1}}} = {{hc} \over E} \times {{\sqrt {2{m_p}E} } \over h}$

$ \Rightarrow {{{\lambda _2}} \over {{\lambda _1}}} = {c \over {\sqrt E }}\sqrt {2{m_p}} $

$ \Rightarrow {{{\lambda _2}} \over {{\lambda _1}}} \propto {E^{ - 1/2}}$

Given, radius of hydrogen atom, r = 2.5 $\times$ 10^$-$13 m

Volume of one atom, $V = {4 \over 3}\pi {r^3}$

Number of atoms in 1 mole = 6.023 $\times$ 10²³

Volume of 1 mole of H-atoms = $N \times {4 \over 3}\pi {r^3}$

$ = 6.023 \times {10^{23}} \times {4 \over 3} \times 3.14 \times {(2.5 \times {10^{ - 13}})^3}$

$ = 3.94 \times {10^{ - 14}}$ m³

According to law of radioactivity, the rate of disintegration is given as

$R = \lambda N = \lambda {N_0}{e^{ - \lambda t}}$

Given, $R = 5000$

$ \Rightarrow 5000 = \lambda {N_0}{e^{ - \lambda t}}$ ...... (i)

After 5 min, $R = 1250$

$\therefore$ $1250 = \lambda {N_0}{e^{ - \lambda (t + 5)}}$ ...... (ii)

On dividing Eq. (i) by Eq. (ii), we get

${{5000} \over {1250}} = {{{e^{ - \lambda t}}} \over {{e^{ - \lambda (t + 5)}}}}$

$ \Rightarrow 4 = {e^{5\lambda }}$

Taking log both sides, we get

${\log _e}4 = {\log _e}({e^{5\lambda }})$

$ \Rightarrow lo{g_e}4 = 5\lambda \Rightarrow \lambda = {1 \over 5}{\log _e}{2^2}$

$ \Rightarrow \lambda = 0.4{\log _e}2$