QJEEAdv08Ph76a2: JEE Advanced 2008 Physics - Class 12 JEEAdv08Ph76a2 Solution: The quantum number n of the state finally populated in He ions…

For H atom,

${E_1} = - {{13.6} \over {{1^2}}} = - 13.6\,eV$

${E_2} = - {{13.6} \over {{2^2}}} = - 3.4\,eV$

Energy released by H atom = E $_2$ $-$ E $_1$

$= - 3.4 - ( - 13.6) = 10.2\,eV$

This energy will be absorbed by He atom. Thus, for He atom

$10.2 = - 13.6 \times {2^2}\left( {{1 \over {{2^2}}} - {1 \over {{n^2}}}} \right)$

$0.1875 = {1 \over 4} - {1 \over {{n^2}}}$

${n^2} = 16$

$n = 4$