JEE Advanced 2021 — Paper 1 — Question 6

For hydrogen atom, $\mathrm{z}=1$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$,

where $\mathrm{R}=1.0973 \times 10^{7} \mathrm{~m}^{-1}=1.1 \times 10^{7} \mathrm{~m}^{-1}=$ Rydberg constant. For the Lyman series, $n_{1}=1$

and $n_{2}=2,3,4, \ldots \ldots \infty$ $\frac{1}{\lambda}=R\left(1-\frac{1}{n_{2}^{2}}\right)$ $\lambda=\lambda_{\text {max }}$,

when $\mathrm{n}_{2}=2$

$\frac{1}{\lambda_{\text {max }}}=\frac{3 \mathrm{R}}{4} \Rightarrow \lambda_{\text {max }}=\frac{4}{3 \mathrm{R}}=121.5 \mathrm{~nm}$

$\lambda=\lambda_{\text {min }}$, when $\mathrm{n}_{2}=\infty$

$\frac{1}{\lambda_{\text {min }}}=\mathrm{R} \Rightarrow \lambda_{\text {min }}=\frac{1}{\mathrm{R}}=91.1 \mathrm{~nm}$ Also, $\lambda=\left(\frac{\mathrm{n}_{2}^{2}}{\mathrm{n}_{2}^{2}-1}\right) \frac{1}{\mathrm{R}}=\left[1+\frac{1}{\left(\mathrm{n}_{2}^{2}-1\right)}\right] \lambda_{0}=\left(1+\frac{1}{\mathrm{~m}^{2}}\right) \lambda_{0}$

$\lambda=\left(1+\frac{1}{\mathrm{~m}^{2}}\right) \lambda_{0}$ where, $\mathrm{m}^{2}=\left(\mathrm{n}_{2}^{2}-1\right)=$ an integer $\mathrm{m}=\sqrt{\mathrm{n}_{2}^{2}-1}=$ not an integer For the Balmer series, $\mathrm{n}_{1}=2$ and $\mathrm{n}_{2}=3,4,5,6$, $\qquad$ ,

$\infty$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ $\lambda=\lambda_{\text {max }}$,

when $\mathrm{n}_{2}=3$ $\frac{1}{\lambda_{\text {max }}}=\frac{5 R}{36}$ $\lambda_{\text {max }}=\frac{36}{5 \mathrm{R}}=656.2 \mathrm{~nm}$ $\lambda=\lambda_{\text {min }}$, when $\mathrm{n}_{2}=\infty$ $\Rightarrow \lambda_{\text {min }}=\frac{4}{\mathrm{R}}=364.5 \mathrm{~nm}$

Hence, for the Balmer series, $\frac{\lambda_{\text {max }}}{\lambda_{\text {min }}}=\frac{36 / 5 R}{4 / R}=\frac{9}{5}$

For the Paschen series, $n_{1}=3$ and $n_{2}=4,5,6, \ldots \ldots \ldots . . \infty$

$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{9}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$

$\lambda=\lambda_{\text {max }}$, when $\mathrm{n}_{2}=4$

$\frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 \mathrm{R}}{144}$

$\Rightarrow \lambda_{\text {max }}=\frac{144}{7 \mathrm{R}}=1874.7 \mathrm{~nm}$

$\lambda=\lambda_{\text {min }}$, when $\mathrm{n}_{2}=\infty$

$\frac{1}{\lambda_{\text {min }}}=\frac{\mathrm{R}}{9} \Rightarrow \lambda_{\text {min }}=\frac{9}{\mathrm{R}}=820.2 \mathrm{~nm}$