Atomic Physics Question 7
Question 7 - 2024 (29 Jan Shift 2)
Hydrogen atom is bombarded with electrons accelerated through a potential different of V, which causes excitation of hydrogen atoms. If the experiment is being formed at $\mathrm{T}=0 \mathrm{~K}$. The minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha}{10} \mathrm{~V}$, where $\alpha=$
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Answer: (121)
Solution:
For minimum potential difference electron has to make transition from $n=3$ to $n=2$ state but first electron has to reach to $\mathrm{n}=3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $n=3$ and $n=1$ state.
$\Delta \mathrm{E}=13.6\left[1-\frac{1}{3^{2}}\right] \mathrm{e}=\mathrm{eV}$
$\frac{13.6 \times 8}{9}=\mathrm{V}$
$\mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V}$
So, $\alpha=121$
