Q. An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg's constant $R=10^5 cm^{-1} $ the frequency in hertz of the emitted radiation will be


Frequency, $v= \frac {c}{λ}=c.R \bigg (\frac {1}{n_1^2}- \frac {1}{n_2^2} \bigg ) $
$\hspace15mm =3 \times 10^8 \times 10^7 \bigg (\frac {1}{2^2}- \frac {1}{4^2} \bigg ) $
$\hspace15mm =\frac {9}{16} \times 10^{15} Hz $

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