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The ratio of the velocity of light and the velocity of electron in the first orbit of a hydrogen atom
$[Given h = 6 .624 \times 10^{- 27} erg - \sec; m = 9 .108 \times 10^{- 28} g, r = 0 .529 \times 10^{- 8} cm .]$

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$137$

$\frac{1}{137}$

$\frac{1}{13700}$

$13700$

answer is A.

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Detailed Solution

\begin{array}{l} v = \frac{h}{2 πm} (or) V_{n} = 2.18 \times 10^{8} \times \frac{Z}{n} cm / \sec \\ = \frac{6 .624 \times 10^{- 27}}{2 \times 3 .14 \times 9 .108 \times 10^{- 28} \times 0 .529 \times 10^{- 8}} \\ = 2 .189 \times 10^{8} cm / \sec \\ \frac{c}{v} = \frac{3 \times 10^{8}}{2 .189 \times 10^{8}} = 137 \end{array}

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