Modern Physics Ques 102
- When a certain photosensitive surface is illuminated with monochromatic light of frequency $v$, the stopping potential for the photocurrent is $-V _0 / 2$. When the surface is illuminated by monochromatic light of frequency $v / 2$, the stopping potential is $-V _0$. The threshold frequency for photoelectric emission is
(Main 2019, 12 Jan II)
(a) $\frac{4}{3}$ $ v$
(b) $2 $ $v$
(c) $\frac{3 v}{2}$
(d) $\frac{5 v}{3}$
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Answer:
Correct Answer: 102.(c)
Solution:
Formula:
- Relation between stopping potential and incident light’s frequency is $e V _0=h f-\varphi _0$. where, $V _0$ is the stopping potential and $\varphi _0$ is the the work function of the photosensitive surface.
So, from given data, we have,
$ -e \frac{V _0}{2}=h v-\varphi _0 \quad …….(i) $
and
$ -e V _0=\frac{h v}{2}-\varphi _0 \quad …….(ii) $
Subtracting Eqs. (i) from (ii), we have
$-e V _0-(-\frac{e V _0}{2}) =\frac{h \nu}{2}-h \nu $
$\Rightarrow \quad -\frac{e V _0}{2}=-\frac{h \nu}{2} $
$\Rightarrow \quad e V _0 =h \nu$
Substituting this in Eq. (i), we get
$ \begin{aligned} & -\frac{e V _0}{2}=e V _0-\varphi _0 \\ \Rightarrow \quad & -(\frac{3}{2} e V _0)=-\varphi _0 \quad \text { or } \frac{3}{2} h \nu=\varphi _0 \end{aligned} $
If threshold frequency is $v _0$ then
$ h v _0=\frac{3}{2} h v \Rightarrow v _0=\frac{3}{2} v $
BETA
