The work functions (W_0) of K, Na, Li, Mg and | vedclass

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $\lambda = 450 \ nm = 450 	imes 10^{-9} \ m$.$E = \frac{6.626 	imes

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $\lambda = 450 \ nm = 450 	imes 10^{-9} \ m$.$E = \frac{6.626 	imes 10^{-34} \ J \cdot s 	imes 3 	imes 10^8 \ m/s}{450 	imes 10^{-9} \ m} \approx 4.417 	imes 10^{-19} \ J$.Converting to $eV$: $E = \frac{4.417 	imes 10^{-19} \ J}{1.602 	imes 10^{-19} \ J/eV} \approx 2.76 \ eV$.Photoelectric effect occurs if the energy of the incident photon is greater than or equal to the work function $(E \ge W_0)$.Metals with $W_0 Metals with $W_0 > 2.76 \ eV$ are $Mg (3.70 \ eV)$ and $Cu (4.80 \ eV)$. These will not undergo the photoelectric effect.The number of such metals is $2$.

The work functions $(W_0)$ of $K, Na, Li, Mg$ and $Cu$ are $2.25, 2.30, 2.42, 3.70$ and $4.80 \ eV$ respectively. How many of these metals do not undergo photoelectric effect when a radiation of wavelength $450 \ nm$ is allowed to fall on them? $(1 \ eV = 1.602 \times 10^{-19} \ J)$

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