QJEEM20Ph7a3d: JEE Main 2020 Physics - Class 12 JEEM20Ph7a3d Solution: In the line spectra of hydrogen atoms, difference between the…

For Lyman series :

Shortest wavelength,

$\lambda$ _s = $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$

longest wavelength,

$\lambda$ _l = $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

Given, $\lambda$ _l - $\lambda$ _s = 304 $\mathop A\limits^0$

$\therefore$ $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$ - $R{.1^2}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$ = 304

$\Rightarrow$ ${1 \over R}$ = 912 $\mathop A\limits^0$

For Paschen series :

Shortest wavelength,

$\lambda$ _s = $R{.1^2}\left( {{1 \over {{3^2}}} - {1 \over {{\infty ^2}}}} \right)$ = ${R \over 9}$

longest wavelength,

$\lambda$ _l = $R{.1^2}\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right)$ = ${{7R} \over {144}}$

$\therefore$ $\lambda$ _l - $\lambda$ _s

= ${{144} \over {7R}} - {9 \over R}$

= $\left( {{{144} \over 7} - 9} \right){1 \over R}$

= $\left( {{{144} \over 7} - 9} \right) \times 912$ = 10553 $\mathop A\limits^0$