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The frequency of the radiation emitted by a hydrogen atom for the transition between \mathrm{n=2} and \mathrm{n=1} states is \mathrm{v_0} . What is the frequency of the radiation emitted by the hydrogen atom for transition between \mathrm{n=4} and \mathrm{n=1} states

Option: 1

\mathrm{\frac{3v_{0}}{2}}


Option: 2

\mathrm{2v_{0}}


Option: 3

\mathrm{4v_{0}}


Option: 4

\mathrm{\frac{5v_{0}}{4}}


Answers (1)

\mathrm{\text { since } \Delta E=h v=R h c\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]} (for hydrogen atom)
\mathrm{v=\operatorname{Rc}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \text {, Here }}
\mathrm{v_0=R C\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \Rightarrow v_0=R C\left[1-\frac{1}{4}\right]}
\mathrm{\therefore v_{0}=\frac{3Rc}{4}}
Again, \mathrm{V^{\prime}=\operatorname{Rc}\left[\frac{1}{1^2}-\frac{1}{4^2}\right]=R C\left[1-\frac{1}{16}\right]=\frac{15}{16}\ R C}
\mathrm{\therefore \frac{v^{\prime}}{v_0}=\frac{15}{16} \times \frac{4}{3}=\frac{5}{4}}
\mathrm{v^{\prime}=\frac{5 v_0}{4}}
 

Posted by

Sumit Saini

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