Modern Physics 1 Question 4
13. In a hydrogen like atom electron makes transition from an energy level with quantum number $n$ to another with quantum number $(n-1)$. If $n \gg 1$, the frequency of radiation emitted is proportional to
(a) $\frac{1}{n}$
(b) $\frac{1}{n^{2}}$
(c) $\frac{1}{n^{4}}$
(d) $\frac{1}{n^{3}}$
(2013 Main)
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Solution:
13.6 \times 4 \times \frac{1}{m^{2}}-\frac{1}{n^{2}}=\frac{h c}{108.5 nm} $$
and in second case,
$$ \begin{aligned} & 13.6 \times 4 \times \frac{1}{1^{2}}-\frac{1}{m^{2}}=\frac{h c}{30.4 nm} \\ \Rightarrow \quad 1-\frac{1}{m^{2}} & =\frac{1240 eV}{30.4 \times 13.6 \times 4} \quad \text { Given, } E=\frac{1240 eV}{\lambda(in nm)} \\ \Rightarrow \quad 1-\frac{1}{m^{2}} & =0.74980 \approx 0.75 \\ \text { or } \quad \frac{1}{m^{2}} & =1-0.75=0.25 \end{aligned} $$
$$ \Rightarrow \quad m^{2}=\frac{1}{0.25}=4 $$
Hence, $m=2$
So, by putting the value of $m$ in Eq. (ii), we get
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