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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