Hydrogen atom is bombarded with electrons accelerated through a potential difference of $\mathrm{V}$, which causes excitation of hydrogen atoms. If the experiment is being performed 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 $\alpha=$ __________.
Answer (integer)
121
Solution
<p>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 $\mathrm{n=3}$ and $\mathrm{n=1}$ state.</p>
<p>$$\begin{aligned}
& \Delta \mathrm{E}=13.6\left[1-\frac{1}{3^2}\right] \mathrm{e}=\mathrm{eV} \\
& \frac{13.6 \times 8}{9}=\mathrm{V} \\
& \mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V}
\end{aligned}$$</p>
<p>So, $\alpha=121$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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