QJEEAdv03Ch68ae: JEE Advanced 2003 Chemistry - Class 11 JEEAdv03Ch68ae Solution: Wavelength of high energy transition of H-atoms is 91.2 nm.…

${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

$\therefore$ ${1 \over {91.2}} = {R_H}\left[ {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right]$

Since R_H = constant

${{{\lambda _{He}}} \over {{\lambda _H}}} = {{Z_H^2} \over {Z_{He}^2}} = {1 \over 4}$

or, ${\lambda _{He}} = {{{\lambda _H}} \over 4} = {{91.2} \over 4} = 22.8$ nm