The work functions of Ag, Mg, K and Na respec | vedclass

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(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $h = 6.626 	imes 10^{-34} \ J \ s$,$c = 3 	imes 10^8 \ m/s$,and $\l

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$3$

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(B) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $h = 6.626 	imes 10^{-34} \ J \ s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 300 	imes 10^{-9} \ m$:$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{300 	imes 10^{-9}} = 6.626 	imes 10^{-19} \ J$.Converting this energy to $eV$: $E = \frac{6.626 	imes 10^{-19}}{1.6022 	imes 10^{-19}} \approx 4.135 \ eV$.Photoelectric effect occurs if the incident photon energy is greater than the work function $(\Phi)$ of the metal.Comparing $\Phi$ values:$Ag: 4.3 \ eV > 4.135 \ eV$ (No emission)$Mg: 3.7 \ eV $K: 2.25 \ eV $Na: 2.30 \ eV Thus,$Mg, K,$ and $Na$ will eject electrons. The total number of such metals is $3$.

The work functions of $Ag, Mg, K$ and $Na$ respectively in $eV$ are $4.3, 3.7, 2.25, 2.30$. When an electromagnetic radiation of wavelength of $300 \ nm$ is allowed to fall on these metal surfaces,the number of metals from which the electrons are ejected is $(eV = 1.6022 \times 10^{-19} \ J)$

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