Question

Explain...................​

Answer 1

$\sf{\Large {\underline {STRUCTURE \: OF\:THE\:ATOM}}}$ :

I. E. ( Ionisation Enthalpy) of H- atom = 1.312 × ${10}^{6}$

$\delta{E}$ = $E_2$ - $E_1$

$\delta{E}$ = - [ $(1.312×{10} ^{6})\: ÷\: ( {2}^{2}$)] - [ - $(1.312×{10} ^{6})\: ÷\: ( {1}^{2}$)]

$\delta{E}$ = - 3.28 × ${10}^{5}$ + 13.12 × ${10}^{5}$

$\delta{E}$ = 9.84 × ${10}^{5}$ J / mol.

$\sf{\underline {So, \:Option\:(\:d\:) \:is\:correct.}}$