QJEEAdv21Ph9ee7: JEE Advanced 2021 Physics - Class 12 JEEAdv21Ph9ee7 Solution: Which of the following statement(s) is(are) correct about the…

For Balmer series

\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{n^2}\right) n>2

$\lambda$ is minimum for transition from $n_1 \rightarrow \infty$ to $n_2=2$ .

$\frac{1}{\lambda_{\min }}=\frac{R}{4}$

$\lambda$ is maximum for transition from $n_1=3$ to $n_2=2$

$\frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36}$

Dividing the above two equations, we get

$\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{R}{4} \times \frac{36}{5 R}=\frac{9}{5}$

Option (A) is correct.

For Lyman series,

$\lambda_{\min }=91.18 \mathrm{~nm}$

$\lambda_{\max }=121.57 \mathrm{~nm}$

For Balmer series,

$\lambda_{\min }=364.60 \mathrm{~nm}$

$\lambda_{\max }=656.30 \mathrm{~nm}$

For Pashen series,

$\lambda_{\min }=820.40 \mathrm{~nm}$

$\lambda_{\max }=1875 \mathrm{~nm}$

We can see from the above that there is no overlap between the wavelength ranges of Balmer and Paschen series. so, option (B) is incorrect.

Similarly wavelength ranges of Lyman and Balmer series also do not overlap. Hence, option (D) is correct.

Shortest wavelength of Lyman series

$\begin{aligned}\frac{1}{\lambda_0} & =R\left(1-\frac{1}{\infty}\right) \\\\\lambda_0 & =\frac{1}{R}\end{aligned}$

For Lyman series, $\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{n^2}\right)$

$\begin{aligned}& =\frac{1}{\lambda_0}\left(1-\frac{1}{n^2}\right) \\\\\lambda & =\frac{\lambda_0}{1-\frac{1}{n^2}}=\frac{\lambda_0 n^2}{n^2-1} \\\\& =\frac{\lambda_0\left(n^2-1\right)+1}{n^2-1} \\\\& =\lambda_0\left(1+\frac{1}{n^2-1}\right)\end{aligned}$

If $n^2-1=m^2$ ( $m$ is an integer)

then this is rarely true as $\left(n^2-1\right)$ cannot always be the square of an integer. Hence, option (C) is incorrect.