Atoms And Nuclei Question 20
Question: An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg’s constant $ R=10^{5}c{{m}^{-1}} $ . The frequency in Hz of the emitted radiation will be [CPMT 1976]
Options:
A) $ \frac{3}{16}\times 10^{5} $
B) $ \frac{3}{16}\times 10^{15} $
C) $ \frac{9}{16}\times 10^{15} $
D) $ \frac{3}{4}\times 10^{15} $
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Answer:
Correct Answer: C
Solution:
$ \frac{1}{\lambda }=R,( \frac{1}{2^{2}}-\frac{1}{4^{2}} )=\frac{3R}{16}\Rightarrow \lambda =\frac{16}{3R}=\frac{16}{3}\times {{10}^{-5}}cm $
Frequency $ n=\frac{c}{\lambda }=\frac{3\times 10^{10}}{\frac{16}{3}\times {{10}^{-5}}}=\frac{9}{16}\times 10^{15}Hz $
BETA
