The correct answer is option (C)

$\begin{array}{rl}& \frac{1}{\lambda }=\mathrm{R}(\frac{1}{(1{)}^2}-\frac{1}{{\mathrm{n}}^2})\mathrm{n}=2,3,4,\dots \dots \\ & {\left(\frac{1}{{\lambda }_{\mathrm{L}}}\right)}_{max}=\mathrm{R}(\frac{1}{(1{)}^2}-\frac{1}{(2{)}^2})(\because \frac{1}{\mathrm{R}}\simeq 912\stackrel{o}{A})\\ & {\left({\lambda }_{\mathrm{L}}\right)}_{max}=\frac{4}{3}\frac{\mathrm{L}}{\mathrm{R}}\\ & {\left({\lambda }_{\mathrm{L}}\right)}_{max}=\frac{4}{3}\times 912\stackrel{o}{A}=4\times 304\stackrel{o}{A}=1216\stackrel{o}{A}\\ & {\left(\frac{1}{{\lambda }_{\mathrm{L}}}\right)}_{min}=\mathrm{R}(\frac{1}{(1{)}^2}-\frac{1}{(\mathrm{\infty }{)}^2})\\ & {\left({\lambda }_{\mathrm{L}}\right)}_{min}=\frac{1}{\mathrm{R}}\simeq 912\stackrel{o}{A}\end{array}$

Range of $\lambda$ is $912\stackrel{o}{A}$ to $1216\stackrel{o}{A}$ which lies in U.V. region.

The correct answer is option (A)

Shortest wavelength in Balmer series when transition of $e^{-}$ from $\mathrm{\infty }$ to $\mathrm{n}=2$

$\because \frac{1}{\lambda }={\mathrm{Rz}}^2[\frac{1}{2^2}-\frac{1}{{\mathrm{\infty }}^2}]$

$\frac{1}{\lambda }=\frac{R}{4}$ .... (1)

Shortest wavelength is Bracket series when transition of $e^{-}$ from $\mathrm{\infty }$ to $\mathrm{n}=4$

$\frac{1}{{\lambda }^{\prime }}=\mathrm{R}(1{)}^2[\frac{1}{4^2}-\frac{1}{{\mathrm{\infty }}^2}]\Rightarrow \frac{1}{{\lambda }^{\prime }}=\frac{\mathrm{R}}{16}$ ..... (2)

Eq. (1) / Eq. (2)

$\frac{{\lambda }^{\prime }}{\lambda }=\frac{\mathrm{R}}{4}\times \frac{16}{\mathrm{R}}\Rightarrow {\lambda }^{\prime }=4\lambda$

The Correct Answer is Option (B)

In case of Balmer series :

n₁ = $\mathrm{\infty }$, n₂ = 2

$\therefore$ $\frac{1}{{\lambda }_B}=Rc\left(\frac{1}{2^2}-\frac{1}{{\mathrm{\infty }}^2}\right)$ = $\frac{Rc}{4}$

In case of Lyman series :

n₁ = $\mathrm{\infty }$, n₂ = 1

$\therefore$ $\frac{1}{{\lambda }_L}=Rc\left(\frac{1}{1^2}-\frac{1}{{\mathrm{\infty }}^2}\right)$ = Rc

$\therefore$ $\frac{{\lambda }_B}{{\lambda }_L}$ = $\frac{4}{1}$ = 4

The Correct Answer is Option (C)

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,

$\frac{1}{\lambda }=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

On jumping from 3^rd orbit to 2^nd orbit,

$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ = $\frac{5R}{36}$

On jumping from 4^th orbit to 3^rd orbit,

$\frac{1}{{\lambda }^{\prime }}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$ = $\frac{7R}{144}$

$\therefore$ $\lambda$' = $\frac{144}{7}\times \frac{5\lambda }{36}$ = $\frac{20\lambda }{7}$

The Correct Answer is Option (A)

The wave number of the last line of the Balmer series in hydrogen spectrum is given by

$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{{\mathrm{\infty }}^2}\right)$

= $\frac{R}{4}$ = $\frac{{10}^7}{4}$ = 0.25 $\times$ 10⁷ m^$-$1

The Correct Answer is Option (B)

For the longest wavelength in the Lyman series,

n₁ = 1 and n₂ = 2

$\frac{1}{{\lambda }_L}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ = $\frac{3R}{4}$

For the longest wavelength in the Balmer series,

n₁ = 2 and n₂ = 3

$\frac{1}{{\lambda }_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ = $\frac{5R}{36}$

$\therefore$ $\frac{{\lambda }_L}{{\lambda }_B}=\frac{\frac{4}{3R}}{\frac{36}{5R}}$ = $\frac{5}{27}$

The Correct Answer is Option ()

Energy of radiation :

E = $\frac{12375}{975}$ = 12.7 eV

After absorbing a photon of energy 12.75 eV, the electron will reach to third excited state of energy –0.85 eV, since energy difference corresponding to n = 1 and n = 4 is 12.75 eV.

$\therefore$ Number of spectral lines emitted

= $\frac{n\left(n-1\right)}{2}=\frac{4\left(4-1\right)}{2}$ = 6

The Correct Answer is Option (C)

We see that wavelength of Lyman series,

n_f = 3, n_i = 4

$\frac{1}{{\lambda }_L}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right]$ = $\frac{7R}{144}$

We see that wavelength of Balmer series :

n_f = 2, n_i = 3

$\frac{1}{{\lambda }_B}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$ = $\frac{5R}{36}$

Now ratio of longest wavelengths corresponds to Lyman and Balmer series:

$\frac{{\lambda }_L}{{\lambda }_B}=\frac{5}{36}\times \frac{144}{7}$ = $\frac{5}{27}$

The Correct Answer is Option (D)

The frequency of the transition v $\propto$ $\frac{1}{n^2}$

when n = 1, 2, 3.