⇒Structure of Atom and Nuclei

Correct Option is (B)

De-Broglie wavelength $\lambda \propto \frac{1}{p}$ where $\mathrm{p}$ is momentum

For an electron velocity $\mathrm{v}\propto \frac{1}{\mathrm{n}}$,

hence $\mathrm{p}\propto \frac{1}{\mathrm{n}}$

$\therefore \frac{{\lambda }_1}{{\lambda }_4}=\frac{1}{4}$

$\therefore {\lambda }_1=\frac{{\lambda }_4}{4}$

Correct Option is (A)

Series limit of Lyman series is given by

$\frac{1}{{\lambda }_1}=\mathrm{R}(\frac{1}{1}-\frac{1}{\mathrm{\infty }})=\mathrm{R}$

Series limit of Balmer series is given by

$\frac{1}{{\lambda }_3}=\mathrm{R}(\frac{1}{4}-\frac{1}{\mathrm{\infty }})=\frac{\mathrm{R}}{4}$

First line of Lyman series is given by

$\begin{array}{rl}& \frac{1}{{\lambda }_2}=\mathrm{R}(\frac{1}{1}-\frac{1}{4})=\mathrm{R}-\frac{\mathrm{R}}{4}=\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_3}\\ \therefore & \frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}\end{array}$

Correct Option is (C)

In the region $0\le x\le 1$, the potential energy of the particle is $E$.

Total energy is $\mathrm{nE}$.

Hence, kinetic energy, $K=nE-E=(n-1)E$

Its momentum, ${\mathrm{p}}_1=\sqrt{2\mathrm{mK}}=\sqrt{2\mathrm{m}(\mathrm{n}-1)\mathrm{E}}$

$\therefore {\lambda }_1=\frac{\mathrm{h}}{{\mathrm{p}}_1}=\frac{\mathrm{h}}{\sqrt{2\mathrm{m}(\mathrm{n}-1)\mathrm{E}}}$

In the region $x>1,\mathrm{PE}$ is zero.

Hence, total energy is kinetic energy.

$\begin{array}{rl}& \therefore \mathrm{K}=\mathrm{nE}\\ & \therefore {\lambda }_2=\frac{\mathrm{h}}{{\mathrm{p}}_2}=\frac{\mathrm{h}}{\sqrt{2\mathrm{mK}}}\\ & \therefore \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}}\end{array}$

Correct Option is (B)

Gyromagnetic ratio = $\frac{\mathrm{e}}{2\mathrm{m}}$

It is independent of the orbit of the electron.

Correct Option is (D)

Initially the electron is in the third excited state i.e. n = 4. Hence we can have transitions :

$4\to 3,4\to 2,4\to 1$

$3\to 2,3\to 1$

$2\to 1$

Hence, six transitions are possible.