An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg's constant R=107m−1the frequency in Hz of there emitted radiation will be

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An electron jumps from the 4^th orbit to 2^ndorbit of hydrogen atom. Given the Rydberg's constant $R = 10^{7} m^{- 1}$ the frequency in Hz of there emitted radiation will be

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$(3 / 16) 10^{5}$

$(16 / 3) 10^{5}$

$(9 / 16) 10^{15}$

$(3 / 4) 10^{15}$

answer is C.

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Detailed Solution

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] v = \frac{c}{λ} = CR [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] = (3 \times 10^{8}) (10^{7}) [\frac{1}{(2)^{2}} - \frac{1}{(4)^{2}}] = (3 \times 10^{8}) (10^{7}) [\frac{1}{(2)^{2}} - \frac{1}{(4)^{2}}] = \frac{9}{16} \times 10^{15} Hz$