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The work functions for metals $A, B$ and $C$ are respectively $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 eV$ . According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength $4100 Å$ is/are
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A and B only
That metal will emit photoelectrons which has work function lower than that obtained with the radiation of $4100 Å$.
Work function for wavelength of $4100 Å$ is
$\begin{aligned} W & =\frac{h c}{\lambda} \\ & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4100 \times 10^{-10}} \\ & =4.8 \times 10^{-19} \mathrm{~J} \\ & =\frac{48 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=3 \mathrm{eV}\end{aligned}$
Now, we have
$\begin{aligned}& W_A=1.92 \mathrm{eV} \\& W_B=2.0 \mathrm{eV} \\
& W_C=5 \mathrm{eV}\end{aligned}$
Since, $W_A \lt W$ and $W_B \lt W$, hence, $A$ and $B$ will emit photoelectrons.
Work function for wavelength of $4100 Å$ is
$\begin{aligned} W & =\frac{h c}{\lambda} \\ & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4100 \times 10^{-10}} \\ & =4.8 \times 10^{-19} \mathrm{~J} \\ & =\frac{48 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=3 \mathrm{eV}\end{aligned}$
Now, we have
$\begin{aligned}& W_A=1.92 \mathrm{eV} \\& W_B=2.0 \mathrm{eV} \\
& W_C=5 \mathrm{eV}\end{aligned}$
Since, $W_A \lt W$ and $W_B \lt W$, hence, $A$ and $B$ will emit photoelectrons.
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