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If an electron in hydrogen atom jumps from an orbit of level \( n=3 \) to an orbit of level \( n=2 \). the
emitted radiation has a frequency ( \( R= \) Rydberg constant, \( C= \) velocity of light)
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emitted radiation has a frequency ( \( R= \) Rydberg constant, \( C= \) velocity of light)
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Verified Answer
The correct answer is:
\( \frac{5 R C}{36} \)
We know that
\[
\begin{array}{l}
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\text { Using } \lambda f=c \Rightarrow \frac{1}{\lambda}=\frac{f}{c} \\
\frac{f}{c}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \Rightarrow f=R c\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\text { Given } n_{1}=2 ; n_{2}=3 \\
\Rightarrow f=R c\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R c\left(\frac{1}{4}-\frac{1}{9}\right)=R c \frac{(9-4)}{4 \times 9}=\frac{5 R C}{36}
\end{array}
\]
Therefore, frequency of emitted radiation is \( \frac{5}{36} \mathrm{Rc} \)
\[
\begin{array}{l}
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\text { Using } \lambda f=c \Rightarrow \frac{1}{\lambda}=\frac{f}{c} \\
\frac{f}{c}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \Rightarrow f=R c\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\text { Given } n_{1}=2 ; n_{2}=3 \\
\Rightarrow f=R c\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R c\left(\frac{1}{4}-\frac{1}{9}\right)=R c \frac{(9-4)}{4 \times 9}=\frac{5 R C}{36}
\end{array}
\]
Therefore, frequency of emitted radiation is \( \frac{5}{36} \mathrm{Rc} \)
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