The expression for the frequency of radiation emitted when a hydrogen atom de-excites from level \( n \) to level \( (n - 1) \), for a large value of \( n \) becomes
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{me^4}{4\varepsilon_0^2 h^3} \times \frac{1}{n^3} \)
In the Bohr model the energy of the \( n \)th level of hydrogen is
\( \displaystyle E_n = -\frac{me^4}{8\varepsilon_0^2 h^2 n^2} \)
The emitted frequency is the energy difference divided by \( h \) (Bohr's frequency condition):
\( \displaystyle \nu = \frac{E_n - E_{n-1}}{h} = \frac{me^4}{8\varepsilon_0^2 h^3}\left[\frac{1}{(n-1)^2} - \frac{1}{n^2}\right] \)
\( \displaystyle \frac{1}{(n-1)^2} - \frac{1}{n^2} = \frac{2n - 1}{n^2 (n-1)^2} \)
For large \( n \), \( 2n - 1 \approx 2n \) and \( (n-1)^2 \approx n^2 \), so this becomes
\( \displaystyle \frac{2n}{n^4} = \frac{2}{n^3} \)
\( \displaystyle \nu = \frac{me^4}{8\varepsilon_0^2 h^3} \times \frac{2}{n^3} = \frac{me^4}{4\varepsilon_0^2 h^3} \times \frac{1}{n^3} \)