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  • The maximum KE of the photoelectron is found to be 6.63\times10^{-19}J, when the metal is irradiated with a radiation of frequency 2\times10

The maximum KE of the photoelectron is found to be 6.63 x10-19 J, when the metal is irradiated with radiation of frequency 2 x 1015 Hz. The threshold frequency of the metal is:

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\\\text{Absorbed energy }= \text{threshold energy }+ KE \\ $nv=nv_{0}+K E$ \\ $\therefore n v_{0}=n v-K E$ \\ $6.26 \times 10^{-34} \times v_{0}=6.26 \times 10^{-34} \times 2 \times 10^{14}-6.63 \times 10^{-20}$ \\ $\therefore v_{0}=9.999 \times 10^{13}$ $\simeq 1 \times 10^{14} s^{-1}$

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