JEE PYQ: Atomic Physics Question 25
Question 25 - 2020 (02 Sep Shift 2)
In a hydrogen atom the electron makes a transition from $(n+1)^{th}$ level to the $n^{th}$ level. If $n » 1$, the frequency of radiation emitted is proportional to:
(1) $\frac{1}{n}$
(2) $\frac{1}{n^3}$
(3) $\frac{1}{n^2}$
(4) $\frac{1}{n^4}$
Show Answer
Answer: (2)
Solution
$\Delta E = Rhc\left[\frac{1}{n^2} - \frac{1}{(n+1)^2}\right] = Rhc \cdot \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = R \cdot c \cdot \frac{1+2n}{n^2(n+1)^2}$. For $n » 1$: $\nu = R \cdot c \cdot \frac{2n}{n^2 \times n^2} = \frac{2RC}{n^3}$. So $\nu \propto \frac{1}{n^3}$.
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