Modern Physics 2 Question 6
6. 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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Solution:
- 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,
$$ E=h v=\varphi+eV _0 $$
(here, $\varphi$ is work function of the metal and $V _0$ is stopping potential)
For $\lambda _1$ wavelength’s wave,
$$ \begin{array}{rlrl} E _1 & =h \nu _1=\varphi+eV _{01} \\ \text { or } \quad \frac{h c}{\lambda _1} & =\varphi+eV _{01} \\ \text { Similarly, } & \frac{h c}{\lambda _2} & =\varphi+eV _{02} \end{array} $$
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,
$$ \begin{gathered} \Delta V=1240 \frac{1}{300}-\frac{1}{400} \frac{nmV}{nm} \\ =1240 \times \frac{1}{1200} V \\ \Rightarrow \quad \Delta V=1.03 . V \approx 1 V \end{gathered} $$
