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Question

Find the frequency of light which ejects electron from a metal surface. Fully stopped by a retarding potential of $$3V$$, the photoelectric effect begins in this metal at a frequency of $$6 \times 10^{14} Hz$$. Find the work function for this metal.

Solution
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According to Einstein's photoelectric equation,
$$E_k = hv - W$$
If $$V_s$$ is retarding or stopping potential and $$v_0$$ the threshold frequency, then above equation becomes
$$eV_s = hv - hv_0$$
or $$hv = eV_s + hv_0$$
or $$v = \dfrac{eV_s}{h} + v_0$$
Hence, $$e = 1.6 \times 10^{-10}$$ coulomb, $$V_s = 3V,$$ and $$v_0 = 6 \times 10^{14} Hz$$
Required frequency
$$v = \dfrac{1.6 \times 10^{-10} \times 3}{6.63 \times 10^{-34} } + 6 \times 10^{14}$$
$$= 7.24 \times 10^{14} + 6 \times 10^{14}$$
$$= 13.24 \times 10^{14} Hz$$
$$= 1.324 \times 10^{15} Hz$$

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