The ratio of $(E_{2} - E_{1})$ to $(E_{4} - E_{3})$ for $H$ - atom is approximately.

10.2

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15.4

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5.6

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12.4

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Solution

The correct option is B $15.4$
As we know, $E \propto - \frac{Z^{2}}{n^{2}}$
$E_{H} = - \frac{13.6}{n^{2}} = \frac{E_{1}}{n^{2}}$
$E_{2} - E_{1} = - \frac{E_{1}}{4} + E_{1} = \frac{3 E_{1}}{4}$ ----- 1
$E_{4} - E_{3} = - \frac{E_{1}}{16} + \frac{E_{1}}{9} = \frac{7 E_{1}}{16 \times 9}$ ----- 2

From the 1st and 2nd equation we have,
$∴ \frac{E_{2} - E_{1}}{E_{4} - E_{3}} = \frac{3 E_{1}}{4} \times \frac{16 \times 9}{7 E_{1}} = 15.4$

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