If, the binding energy of electrons in a metal is $250 \mathrm{~kJ} / \mathrm{mol}$, what should be the threshold frequency of the striking photons in order to free an electron from the metal surface?

Question

If, the binding energy of electrons in a metal is $250 \mathrm{~kJ} / \mathrm{mol}$, what should be the threshold frequency of the striking photons in order to free an electron from the metal surface?, $6.26 \times 10^{14} \mathrm{~s}^{-1}$, $12.4 \times 10^{14} \mathrm{~s}^{-1}$, $6.26 \times 10^{12} \mathrm{~s}^{-1}$, $12.4 \times 10^{12} \mathrm{~s}^{-1}$

MARKS App · Accepted Answer

Binding energy of 1 mole of electrons $=250 \mathrm{~kJ}$ Binding energy of 1 electron $\begin{aligned} & =(250) /\left(6.022 \times 10^{23}\right) \mathrm{kJ} \ & =4.15 \times 10^{-22} \mathrm{~kJ} \ & =4.15 \times 10^{-19} \mathrm{~J} \end{aligned}$ Threshold energy $\left(h v_0\right)=$ Binding energy $\begin{aligned} \therefore h v_0 & =4.15 \times 10^{-19} \mathrm{~J} \ \text {or } v_0 & =\left(4.15 \times 10^{-19} \mathrm{~J}\right) / h \ & =\left(4.15 \times 10^{19} \mathrm{~J}\right)\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \ & =6.26 \times 10^{14} \mathrm{~s}^{-1} \end{aligned}$ Hence, the correct option is (a).

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