$\begin{array}{1 1}(a)\;96.4KJ&(b)\;97.32KJ\\(c)\;98.19KJ&(d)\;95.63KJ\end{array}$

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Mole of $H_2$ present in one litre =$\large\frac{PV}{RT}$

$\Rightarrow \large\frac{1\times 1}{0.0821\times 298}$

$\Rightarrow 0.0409$

Thus energy needed to brak H-H bonds in 0.0409mole of $H_2=0.0409\times 436=17.83KJ$

Also energy needed to excite on H atom from $1^{st}$ to $2^{nd}$ energy level=$13.6(1-\large\frac{1}{4})$$eV$

$\Rightarrow 10.2eV$

$\Rightarrow 10.2\times 1.6\times 10^{-19}J $

$\therefore$ Energy needed to excite $0.0409\times 2\times 6.02\times 10^{23}$ atoms of H=$10.2\times 1.6\times 10^{-19}\times 0.0409\times 2\times 6.02\times 10^{23}J$

$\Rightarrow 80.36KJ$

Total energy needed=$17.83+80.36$

$\Rightarrow 98.19KJ$

Hence (c) is the correct answer.

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