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Q.

An electron jumps from the 4^th orbit to the 2^nd orbit of hydrogen atom. Given : the Rydberg’s constant R=10⁵cm^–1. The frequency in Hz, of the emitted radiation will be

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a

$\frac{9}{16} \times 10^{5}$

b

$\frac{3}{6} \times 10^{5}$

c

$\frac{9}{16} \times 10^{15}$

d

$\frac{3}{16} \times 10^{5}$

answer is D.

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Detailed Solution

$\begin{array}{l} ΔE = {RchZ}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ ∴ hv = {RchZ}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ ∴ v = 10^{7} \times 3 \times 10^{8} (1)^{2} (\frac{1}{2^{2}} - \frac{1}{4^{2}}) \\ ∴ v = \frac{9}{16} \times 10^{15} HZ \end{array}$

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An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given : the Rydberg’s constant R=105cm–1. The frequency in Hz, of the emitted radiation will be