Electrons in a sample of gas containing hydrogen-like atom (Z = 3) are in fourth excited state. When photons emitted only due to transition from third excited state to second excited state are incident on a metal plate photo-electrons are ejected. The stopping potential for these photo-electrons is 3.95 V. Now, if only photons emitted due to transition from fourth excited state to third excited state are incident on the same metal plate, the stopping potential for the emitted photo-electrons will be approximately equal to

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0.65 eV

0.85 eV

0.75 eV

None of these

answer is B.

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Detailed Solution

$\begin{array}{l} hv = 13 .6 {(3)}^{2} [\frac{1}{4^{2}} - \frac{1}{5^{2}}] = 2 .75 eV \\ for n = 4 to n = 3 \\ hv = (13 .6) \times {(3)}^{2} [\frac{1}{3^{2}} - \frac{1}{4^{2}}] = 5 .95 eV \\ for shorter wavelength \\ 3 .95 = 5 .95 - ϕ \Rightarrow ϕ = 2 eV \\ for longer wavelength {eV}_{s} = 2 .75 - 2 = 0 .75 eV \end{array}$