If the maximum kinetic energy of photoelectrons ejected from a metal surface when it is irradiated with a…
If the maximum kinetic energy of photoelectrons ejected from a metal surface when it is irradiated with a radiation of frequency $4 \times 10^{14} \mathrm{~s}^{-1}$ is $6.63 \times 10^{-20} \mathrm{~J}$, then the threshold frequency of the metal is
$2 \times 10^{14} \mathrm{~s}^{-1}$
$1 \times 10^{14} \mathrm{~s}^{-1}$
$3 \times 10^{14} \mathrm{~s}^{-1}$
$1 \times 10^{-14} \mathrm{~s}^{-1}$
Solution
The threshold energy for production of a particle is the minimum K.E., a pair of travelling particles must have when they collide.
$
\begin{aligned}
\text { Absorbed energy } & =\mathrm{KE}+\text { threshold energy } \\
h v^{\prime} & =\mathrm{KE}+h v_0^{\prime} \\
h v_0^{\prime} & =h v^{\prime}-\mathrm{KE}
\end{aligned}
$
$
\begin{aligned}
& 6.26 \times 10^{-34} \times v_0 \\
& =6.26 \times 10^{-34} \times 4 \times 10^{14}-6.63 \times 10^{-20} \mathrm{~J} \\
& \therefore \quad v_0=2.999 \times 10^{13} \mathrm{~s}^{-1}=3 \times 10^{14} \mathrm{~s}^{-1} \\
&
\end{aligned}
$
Hence, threshold frequency $\left(v_0^{\prime}\right)$ of metal is
$
3 \times 10^{14} \mathrm{~s}^{-1}
$