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(D) The energy of the incident radiation is calculated using the formula $E = \frac{hc}{\lambda}$.For $\lambda = 300 \ nm = 300 	imes 10^{-9} \ m$,$E

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$3$

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(D) The energy of the incident radiation is calculated using the formula $E = \frac{hc}{\lambda}$.For $\lambda = 300 \ nm = 300 	imes 10^{-9} \ m$,$E = \frac{6.626 	imes 10^{-34} \ J \cdot s 	imes 3 	imes 10^8 \ m/s}{300 	imes 10^{-9} \ m} \approx 6.626 	imes 10^{-19} \ J \approx 4.14 \ eV$.The work functions $(\Phi)$ of the metals are approximately: $Li \approx 2.5 \ eV$,$Mg \approx 3.7 \ eV$,$Ag \approx 4.3 \ eV$,$Cu \approx 4.7 \ eV$,and $K \approx 2.3 \ eV$.Photoelectric effect occurs when $E > \Phi$.Comparing the values,$Li, Mg,$ and $K$ have work functions less than $4.14 \ eV$.Thus,$3$ metals show the photoelectric effect.

When radiation of $300 \ nm$ is shone on five metals,namely $Li, Mg, Ag, Cu,$ and $K$,the number of metals that show the photoelectric effect is:

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