# 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

AFMCAFMC 2010

Solution:

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

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