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

Solution:

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 $

You must select option to get answer and solution

Questions from AFMC 2010

Physics Most Viewed Questions

9. The spectrum of an oil flame is an example for ...........

KCET 2010 Dual Nature Of Radiation And Matter