Values of work function (W_0) for a few metal | vedclass

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(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.6 	imes 10^{-34} \ J \ s 	imes 3

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$3$

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(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.6 	imes 10^{-34} \ J \ s 	imes 3 	imes 10^8 \ m \ s^{-1}}{400 	imes 10^{-9} \ m} = 4.95 	imes 10^{-19} \ J$.Converting this energy into electron volts $(eV)$: $E = \frac{4.95 	imes 10^{-19} \ J}{1.6 	imes 10^{-19} \ J/eV} \approx 3.1 \ eV$.Photoelectric effect occurs when the energy of the incident photon is greater than the work function $(W_0)$ of the metal $(E > W_0)$.Comparing $3.1 \ eV$ with the given work functions:$Li (2.42 \ eV) $Na (2.3 \ eV) $K (2.25 \ eV) $Mg (3.7 \ eV) > 3.1 \ eV$ (No)$Cu (4.8 \ eV) > 3.1 \ eV$ (No)$Ag (4.3 \ eV) > 3.1 \ eV$ (No)Thus,$3$ metals $(Li, Na, K)$ will show the photoelectric effect.

Metal	$Li$	$Na$	$K$	$Mg$	$Cu$	$Ag$
$W_0 / eV$	$2.42$	$2.3$	$2.25$	$3.7$	$4.8$	$4.3$

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