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The work function of a metal is $2 \mathrm{eV}$. If a radiation of wavelength $3000 Å$ is incident on it, the maximum kinetic energy of the emitted photoelectrons is (Planck's constant $h=6.6 \times 10^{-34} \mathrm{Js}$; velocity of light $\left.c=3 \times 10^8 \mathrm{~m} / \mathrm{s} ; 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$
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Verified Answer
The correct answer is:
$3.4 \times 10^{-19} \mathrm{~J}$
$$
\text { Maximum } \mathrm{KE}=\frac{h \mathrm{c}}{\lambda}-\phi_0
$$
Given, $\lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m}$,
$$
h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}, \mathrm{C}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \phi=2 \mathrm{eV}
$$
Maximum KE
$$
\begin{aligned}
= & \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} \times \frac{1}{1.6 \times 10^{-12}}-2 \\
= & 4.13-2=2.13 \mathrm{eV}
\end{aligned}
$$
In Joules : $2.13 \times 1.6 \times 10^{-19}=3.41 \times 10^{-19} \mathrm{~J}$
\text { Maximum } \mathrm{KE}=\frac{h \mathrm{c}}{\lambda}-\phi_0
$$
Given, $\lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m}$,
$$
h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}, \mathrm{C}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \phi=2 \mathrm{eV}
$$
Maximum KE
$$
\begin{aligned}
= & \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3000 \times 10^{-10}} \times \frac{1}{1.6 \times 10^{-12}}-2 \\
= & 4.13-2=2.13 \mathrm{eV}
\end{aligned}
$$
In Joules : $2.13 \times 1.6 \times 10^{-19}=3.41 \times 10^{-19} \mathrm{~J}$
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