⇒Structure of Atom and Nuclei

Correct Option is (C)

For spectral series, $\frac{1}{\lambda }=R{Z}^2(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2})$

For the Balmer series, ${\mathrm{n}}_1=2$

The wavelength for $2^{\text{nd }}$ line of the Balmer series is

$\begin{array}{rl}& \frac{1}{{\lambda }_1}={\mathrm{RZ}}^2(\frac{1}{2^2}-\frac{1}{4^2})\\ & \frac{1}{{\lambda }_1}={\mathrm{RZ}}^2(\frac{1}{4}-\frac{1}{16})\\ & \frac{1}{{\lambda }_1}={\mathrm{RZ}}^2\left(\frac{3}{16}\right)\Rightarrow {\lambda }_1=\left[\frac{16}{3}\right]\end{array}$

For the Paschen series, ${\mathrm{n}}_1=3$

The wavelength for $1^{\text{st }}$ line of the Paschen series is

$\begin{array}{rl}\frac{1}{{\lambda }_2}& =R{Z}^2(\frac{1}{3^2}-\frac{1}{4^2})\\ \frac{1}{{\lambda }_2}& =R{Z}^2(\frac{1}{9}-\frac{1}{16})\\ {\lambda }_2& =\frac{144}{7}\\ \frac{1}{{\lambda }_2}& =R{Z}^2\left(\frac{7}{144}\right)\\ \therefore \frac{{\lambda }_1}{{\lambda }_2}& =\frac{16}{3}\times \frac{7}{144}=\frac{7}{27}\end{array}$

Correct Option is (D)

According to Rydberg's formula,

$\frac{1}{\lambda }=\mathrm{R}(\frac{1}{{\mathrm{n}}^2}-\frac{1}{{\mathrm{m}}^2})$

For series limit of Lyman series,

$\begin{array}{rl}& \mathrm{n}=1,\mathrm{m}=\mathrm{\infty },\lambda ={\lambda }_1\\ \therefore & \frac{1}{{\lambda }_1}=\mathrm{R}\end{array}$

For $1^{\text{st }}$ line of Lyman series,

$\begin{array}{rl}& \mathrm{n}=1,\mathrm{m}=2,\lambda ={\lambda }_2\\ \therefore & \frac{1}{{\lambda }_2}=\frac{3\mathrm{R}}{4}\end{array}$

For series limit of Balmer series,

$\begin{array}{rl}& \mathrm{n}=2,\mathrm{m}=\mathrm{\infty },\lambda ={\lambda }_3\\ \therefore & \frac{1}{{\lambda }_3}=\frac{\mathrm{R}}{4}\end{array}$

Now, $\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=\mathrm{R}-\frac{3\mathrm{R}}{4}=\frac{\mathrm{R}}{4}$

$\therefore \frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$

Correct Option is (D)

According to Rydberg's formula,

$\frac{1}{\lambda }=\mathrm{R}(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2})$ ...... (i)

When electron jumps from $2^{\text{nd }}$ exited state to first exited state,

${\mathrm{n}}_2=3,{\mathrm{n}}_1=2,\lambda ={\lambda }_0$, we get

$\frac{1}{{\lambda }_0}=\mathrm{R}(\frac{1}{2^2}-\frac{1}{3^2})$

When electron jumps from $3^{\text{rd }}$ exited state to $2^{\text{nd }}$ orbit,

${\mathrm{n}}_2=4,{\mathrm{n}}_1=2$, we get

$\frac{1}{\lambda }=\mathrm{R}(\frac{1}{4^2}-\frac{1}{2^2})$

$\begin{array}{rl}\therefore \frac{\lambda }{{\lambda }_0}& =\frac{\mathrm{R}(\frac{1}{2^2}-\frac{1}{3^2})}{\mathrm{R}(\frac{1}{2^2}-\frac{1}{4^2})}\\ & =\frac{5}{36}\times \frac{16}{3}=\frac{20}{27}\\ \therefore \lambda & =\frac{20}{27}{\lambda }_0\\ \Rightarrow \mathrm{x}& =27\end{array}$

Correct Option is (D)

According to Bohr's theory of hydrogen atom, the equation for total energy of the electron in the ${\mathrm{n}}^{\text{th }}$ stationary orbit is,

$\begin{array}{rl}& {\mathrm{E}}_{\mathrm{n}}=\frac{-m{\mathrm{Z}}^2{\mathrm{e}}^4}{8{\epsilon }_0{}^2{\mathrm{h}}^2{\mathrm{n}}^2}\\ \therefore & {\mathrm{E}}_{\mathrm{n}}\propto \frac{1}{{\mathrm{n}}^2}\end{array}$

Correct Option is (A)

We know,

$\begin{array}{rl}\mathrm{mvr}& =\frac{\mathrm{nh}}{2\pi }\\ \therefore \mathrm{vr}& =\frac{\mathrm{nh}}{2\pi \mathrm{m}}\text{.... (i)}\end{array}$

Also,

$\begin{array}{rl}& \mathrm{qvB}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}\\ & \therefore \mathrm{mv}=\mathrm{qBr}\text{.... (ii)}\\ & {\mathrm{mv}}^2\mathrm{r}=\mathrm{qBr}\times \frac{\mathrm{nh}}{2\pi \mathrm{m}}\text{....(Multiplying (i) with (ii))}\\ & \mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2=\mathrm{n}\left[\frac{\mathrm{qBh}}{4\pi \mathrm{m}}\right]\end{array}$