$(a)\; n=6,z = 3\qquad(b)\;n=3,z=6\qquad(c)\;n=2,z=5\qquad(d)\;n=4,n=2$

Total energy liberated during transition of electron from $n^{th}$ shell to first excited state

(i.e $2^{nd}$ shell) = 10.20+17.0 = 27.20 eV

= $27.20\times1.602\times10^{-12}$ erg

$\large\frac{hc}{\lambda}$= $R_H \times Z^2hc[\large\frac{1}{2^2}-\large\frac{1}{n^2}]$

$27.20\times1.602\times10^{-12}$ = $R_H \times Z^2hc[\large\frac{1}{2^2}-\large\frac{1}{n^2}]$..... (1)

Similarly total energy liberated during transition of electrons from $n^{th}$ shell to second excited state

(i.e $3^{rd}$ shell ) = 4.25+5.95 = 10.20 eV

= $10.20\times1.602\times10^{-12}$ erg

$\therefore10.20\times1.602\times10^{-12}$ = $R_H \times Z^2hc[\large\frac{1}{3^2}-\large\frac{1}{n^2}]$..... (2)

Dividing Eq (1) by Eq (2)

We get n = 6

On substituting the value of n in equation (1) or (2) we get z=3

Hence answer is (a)

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