Electrons from n=2 to n=1 in Hydrogen atom is made to fall on a metal surface with work function 1.2 eV. The maximum velocity of photo electrons emitted is nearly equal to:

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18 × 10⁵ m/s

2 × 10⁵ m/s

3 × 10⁵ m/s

6 × 10⁵ m/s

answer is D.

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Detailed Solution

KE_max = E – W_o
$\begin{array}{l} ∴ \frac{1}{2} {mV}_{max}^{2} = 13 .6 z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) - W_{o} \\ ∴ \frac{1}{2} \times 9 \times 10^{- 31} {V_{max}}^{2} = 13 .6 (1)^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) - 1 .2 eV \\ ∴ \frac{1}{2} \times 9 \times 10^{- 31} {V_{max}}^{2} = [10 .2 - 1 .2] eV \\ ∴ \frac{1}{2} \times 9 \times 10^{- 31} {V_{max}}^{2} = 9 \times 1 .6 \times 10^{- 19} \\ ∴ V_{max} = 18 \times 10^{5} {ms}^{- 1} \end{array}$

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