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 (α)/(10) V, where α = __________. Numerical answer type. Work it out, then open the answer.

Answer: 121. Balmer lines need the electron to reach at least n = 3 (then 3 → 2 is emitted). At T = 0 K all atoms are in n = 1, so the energy needed is E₃ - E₁ = 13.6(1 - 1/9) = 12.09 eV. Minimum accelerating voltage V ≈ 12.1 V = 121/10 V. So α = 121.

Hydrogen atom is bombarded with electrons accelerated through a… | JEE Main 2024 (29 Jan, Shift 2)

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Transitions, Photon Emission & Excitation

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 $T = 0\text{ K}$ , the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha}{10}\text{ V}$ , where $\alpha =$ __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 121

Balmer lines need the electron to reach at least $n = 3$ (then $3 \to 2$ is emitted).
At $T = 0$ K all atoms are in $n = 1$ , so the energy needed is $E_3 - E_1 = 13.6\left(1 - \frac{1}{9}\right) = 12.09\text{ eV}$ .
Minimum accelerating voltage $V \approx 12.1\text{ V} = \frac{121}{10}\text{ V}$ .
So $\alpha = 121$ .

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