The work function (W_0) of Li, K, Mg, Ag and | vedclass

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(C) For the photoelectric effect to occur,the energy of the incident photon $(E)$ must be greater than or equal to the work function $(W_0)$ of the me

Vedclass · Accepted Answer

$1$

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(C) For the photoelectric effect to occur,the energy of the incident photon $(E)$ must be greater than or equal to the work function $(W_0)$ of the metal,i.e.,$E \ge W_0$.Energy of incident photon $(E) = \frac{hc}{\lambda}$.Given $\lambda = 540 \ nm = 540 	imes 10^{-9} \ m$,$h = 6.626 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$.$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{540 	imes 10^{-9}} \ J = \frac{19.878 	imes 10^{-26}}{540 	imes 10^{-9}} \ J = 0.0368 	imes 10^{-17} \ J = 3.68 	imes 10^{-19} \ J$.Converting $E$ to $eV$: $E = \frac{3.68 	imes 10^{-19}}{1.602 	imes 10^{-19}} \ eV \approx 2.297 \ eV$.Photoelectric effect occurs if $E \ge W_0$.Comparing $E = 2.297 \ eV$ with the work functions:$Li (2.42 \ eV) > 2.297 \ eV$ (No)$K (2.25 \ eV) $Mg (3.70 \ eV) > 2.297 \ eV$ (No)$Ag (4.30 \ eV) > 2.297 \ eV$ (No)$Cu (4.80 \ eV) > 2.297 \ eV$ (No)Only $K$ undergoes the photoelectric effect. Thus,the number of metals is $1$.

The work function $(W_0)$ of $Li, K, Mg, Ag$ and $Cu$ are $2.42, 2.25, 3.70, 4.30$ and $4.80 \ eV$ respectively. The number of metals which undergo photoelectric effect if a radiation of wavelength $540 \ nm$ falls on them is $(1 \ eV = 1.602 \times 10^{-19} \ J)$.

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