Modern Physics Ques 103
- In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 $ $nm$ to $400 $ $nm$. The decrease in the stopping potential is close to $(\frac{h c}{e}=1240)$ $ nmV$
(Main 2019, 11 Jan II)
(a) $0.5 $ $V$
(b) $2.0 $ $V$
(c) $1.5 $ $V$
(d) $1.0 $ $V$
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Answer:
Correct Answer: 103.(d)
Solution:
Formula:
- Given, $\lambda _1=300 $ $ nm$
$ \begin{aligned} \lambda _2 & =400 nm \\ \frac{h c}{e} & =1240 nm \end{aligned} $
Using Einstein equation for photoelectric effect,
$ E=h v=\phi+eV _0 \quad …….(i) $
(here, $\phi$ is work function of the metal and $V _0$ is stopping potential)
For $\lambda _1$ wavelength’s wave,
$E _1 =h \nu _1=\phi+eV _{01} $
$\text { or } \quad \frac{h c}{\lambda _1} =\phi+eV _{01} \quad …….(ii) $
$\text { Similarly, } \frac{h c}{\lambda _2} =\phi+eV _{02} \quad …….(iii)$
From Eqs. (ii) and (iii), we get
$ h c [\frac{1}{\lambda _1}-\frac{1}{\lambda _2}]=e\left(V _{01}-V _{02}\right) \text { or } \frac{h c}{e} [\frac{1}{\lambda _1}-\frac{1}{\lambda _2}]=\Delta V $
By using given values,
$\Delta V=1240 [\frac{1}{300}-\frac{1}{400}] \frac{nmV}{nm} $
$= \quad 1240 \times \frac{1}{1200} V $
$\Rightarrow \quad \Delta V=1.03 . V \approx 1 V$
BETA
