V (stopping potential) is plotted against 1 , where is wavelength… - Vidyadip

MCQ

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$, where $\lambda$ is wavelength of incident radiations, for two metals

A
Metal $1$ may be gold and metal $2$ may be cesium
B
$\theta_1 > \theta_2$, if metal $-1$ is gold and metal- $2$ is cesium
✓
$\theta_1=\theta_2$, for any two metals
D
$\theta_1 > \theta_2$, if metal $-1$ and metal- $2$ are gold and copper respectively

✓

Answer

Correct option: C.

$\theta_1=\theta_2$, for any two metals

c
(c)

$\frac{h c}{\lambda}=\frac{h c}{632.2 \times 10^{-9}}$

$\frac{h c}{\lambda}=e V+w_0$

$e V=\frac{h c}{\lambda}-w_0$

$V=\frac{h c}{e} \times \frac{1}{\lambda}-\frac{w_0}{e}$

$\tan \theta=\frac{h c}{e}=$ constant [Slope of $V$ vs $\frac{1}{\lambda}$ graph]

$\therefore \theta$ is same for all metals.

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