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Q.
The frequency and intensity of the incident beam of light falling on the surface of photoelectric material is increased by a factor of two. This will
Dual Nature of Radiation and Matter
Solution:
As the intensity of the beam is doubled, the no. of photons in the beam is doubled. As a consequence, the photoelectrons emitted will also be doubled. Thus, the photoelectric current increases by a factor of two.
To find the energy of photoelectrons, let initial frequency of beam be $v$ and threshold frequency be $v _0$.
Then, by Einsteins eqn
$
\begin{array}{l}
hv = hv _0+\frac{ mv _{ e }^2}{2} \\
\Rightarrow \frac{ mv _{ e }^2}{2}= hv - hv \_ o \$ \$
\end{array}
$
As the frequency is doubled
$
\begin{array}{l}
2 hv = hv _0+\frac{ m \left( v _{ e }\right)^2}{2} \\
\Rightarrow \frac{ m \left( v ^1 e \right)^2}{2}=2 hv - hv _0
\end{array}
$
As, $2 hv - hv _0>2 hv -2 hv _0$
$
\Rightarrow \frac{ m \left( ve ^1\right)^2}{2}>2\left[\frac{ m ( ve )^2}{2}\right]
$
So, the maximum energy of photoelectrons increases by a factor greater than two.