QJEEM24Phfb37: JEE Main 2024 Physics - Class 12 JEEM24Phfb37 Solution: Hydrogen atom is bombarded with electrons accelerated through a …

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QJEEM24Phfb37: JEE Main 2024 Physics - Class 12 JEEM24Phfb37 Solution: Hydrogen atom is bombarded with electrons accelerated through a …

Hydrogen atom is bombarded with electrons accelerated through a potential difference of $V$ , which causes excitation of hydrogen atoms. If the experiment is being performed at $T = 0 K$ , the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{α}{10} V$ , where $α =$ __________.

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Accepted Answer

For minimum potential difference electron has to make transition from $n = 3$ to $n = 2$ state but first electron has to reach to $n = 3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $n = 3$ and $n = 1$ state.

$\begin{aligned} Δ E = 13.6 [1 - \frac{1}{3^{2}}] e = eV \\ \frac{13.6 \times 8}{9} = V \\ V = 12.09 V \approx 12.1 V \end{aligned}$

So, $α = 121$

QJEEM24Phfb37: JEE Main 2024 Physics - Class 12 JEEM24Phfb37 Solution: Hydrogen atom is bombarded with electrons accelerated through a …

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