⇒Structure of Atom and Nuclei

Correct Option is (C)

$\therefore$ Orbital magnetic moment:

$M_0=\frac{-e}{2m_e}L$

$\therefore$ The orbital magnetic moment is directly proportional to angular momentum L of the electron

Correct Option is (B)

Velocity of electron in the ${\mathrm{n}}^{\text{th }}$ orbit is

$\begin{array}{rl}& {\mathrm{v}}_{\mathrm{n}}=\frac{{\mathrm{e}}^2}{2{\epsilon }_0\mathrm{nh}}\\ & \Rightarrow {\mathrm{v}}_{\mathrm{n}}\propto \frac{1}{\mathrm{n}}\end{array}$

Taking the ratio,

$\therefore \frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}=\frac{1}{2}\Rightarrow {\mathrm{v}}_2=\frac{{\mathrm{v}}_1}{2}$

Change in velocity,

$\begin{array}{rl}\mathrm{\Delta }\mathrm{v}& =|{\mathrm{v}}_2-{\mathrm{v}}_1|\\ & =|\frac{{\mathrm{v}}_1}{2}-{\mathrm{v}}_1|=0.5{\mathrm{v}}_1\end{array}$

Therefore, change in percentage is $50\mathrm{\%}$

Correct Option is (B)

Velocity of electron in the ${\mathrm{n}}^{\text{th }}$ orbit ${\mathrm{V}}_{\mathrm{n}}=\frac{{\mathrm{Ze}}^2}{2{\epsilon }_0\mathrm{nh}}$

$\begin{array}{ll}\Rightarrow & {\mathrm{v}}_{\mathrm{n}}\propto \frac{1}{\mathrm{n}}\\ & \text{ given }{\mathrm{n}}_1=1\text{ and }{\mathrm{n}}_2=2,\\ \therefore & \frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\frac{2}{1}\end{array}$

Correct Option is (A)

Using Rydberg's formula,

$\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{n^2}-\frac{1}{m^2}]$

where $R_H$ is the Rydberg's constant

For calculating the shortest wavelength in the Balmer series of hydrogen atom, $\mathrm{n}=2,\mathrm{m}=\mathrm{\infty }$ and $\mathrm{Z}=1$

$\therefore {\lambda }_1=\frac{4}{{\mathrm{R}}_{\mathrm{H}}}$

For hydrogen like atom, the shortest wavelength is given by,

In Brackett series

$\begin{array}{rl}& \mathrm{n}=4,\mathrm{m}=\mathrm{\infty }\\ \therefore & \frac{1}{{\lambda }_2}={\mathrm{R}}_{\mathrm{H}}\cdot {\mathrm{Z}}^2\left(\frac{1}{16}\right)\\ \therefore {\lambda }_2& =\frac{16}{{\mathrm{R}}_{\mathrm{H}}\cdot {\mathrm{Z}}^2}\end{array}$

Given: ${\lambda }_1={\lambda }_2$,

$\begin{array}{rl}& \therefore \frac{4}{R_H}=\frac{16}{R_H\cdot Z^2}\Rightarrow Z^2=4\\ & \therefore Z=2\end{array}$

Correct Option is (A)

For Paschen series, Longest wavelength corresponds to

$\begin{array}{rl}& {\mathrm{n}}_1=3,{\mathrm{n}}_2={\mathrm{n}}_1+1=4\\ & {\lambda }_{max}=\frac{144}{7\mathrm{R}}\end{array}$

Shortest wavelength corresponds to ${\mathrm{n}}_1=3,{\mathrm{n}}_2=\mathrm{\infty }$

$\begin{array}{rl}{\lambda }_{min}& =\frac{9}{R}\\ \therefore \frac{{\lambda }_{max}}{{\lambda }_{min}}& =\frac{\left(\frac{144}{7R}\right)}{\left(\frac{9}{R}\right)}=\frac{144}{63}\end{array}$