The work functions for metals A, B and C are $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$…
The work functions for metals A, B and C are $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$ respectively the metal (s) which will emit photoelectrons for incident radiation of wavelength $4100 \mathrm{~A}$ is /are $\left[\mathrm{h}=6.63 \times 10^{-19} \mathrm{~J}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right]$
Only C
B and C
Only A
$A$ and $B$
Solution
Energy of electron with an associated wavelength of $4100 Å$ is:
$\Rightarrow \frac{\mathrm{hc}}{\lambda}=4.845 \times 10^{-19} \mathrm{~J}=3.024 \mathrm{eV}$
This incident electron would emit photon from metal whose work potential is less than its energy.
Thus, it would emit photons from metals A and B.
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