Atoms And Nuclei Question 124
Question: When the electron in the hydrogen atom jumps from 2nd orbit to 1st orbit, the wavelength of radiation emitted is l. When the electrons jump from 3rd orbit to 1st orbit, the wavelength of emitted radiation would be [MP PMT 2002]
Options:
A) $ \frac{27}{32}\lambda $
B) $ \frac{32}{27}\lambda $
C) $ \frac{2}{3}\lambda $
D) $ \frac{3}{2}\lambda $
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Answer:
Correct Answer: A
Solution:
$ \frac{1}{\lambda }=R[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ] $
First condition $ \frac{1}{\lambda }=R[ \frac{1}{1^{2}}-\frac{1}{2^{2}} ]\Rightarrow R=\frac{4}{3\lambda } $
Second condition $ \frac{1}{\lambda ‘}=R[ \frac{1}{1^{2}}-\frac{1}{3^{2}} ] $
$ \Rightarrow \lambda ‘=\frac{9}{8R}\Rightarrow \lambda ‘=\frac{9}{8\times \frac{4}{3\lambda }}=\frac{27\lambda }{32} $
BETA
