Modern Physics Ques 74
- In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $(\lambda)$ of incident light and the corresponding stopping potential $\left(V _0\right)$ are given below:
(2016 Adv.)
| $\lambda(\mu m)$ | $\boldsymbol{V} _{\mathbf{0}}$ (Volt) |
|---|---|
| 0.3 | 2.0 |
| 0.4 | 1.0 |
| 0.5 | 0.4 |
Given that $c=3 \times 10^{8} $ $ms^{-1}$ and $e=1.6 \times 10^{-19} C$, Planck’s constant (in units of J-s) found from such an experiment is)
(a) $6.0 \times 10^{-34}$
(c) $6.6 \times 10^{-34}$
(b) $6.4 \times 10^{-34}$
(d) $6.8 \times 10^{-34}$
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Answer:
Correct Answer: 74.(b)
Solution:
Formula:
$ \begin{aligned} \frac{h c}{\lambda}-\phi & =e V _0 \\ \frac{h c}{0.3 \times 10^{-6}}-\phi & =2 e \quad …….(i) \\ \frac{h c}{0.4 \times 10^{-6}}-\phi & =1 e \quad …….(ii) \end{aligned} $
Subtracting Eq. (ii) from Eq. (i)
$h c (\frac{1}{0.3}-\frac{1}{0.4}) 10^{6}=e $
$h c (\frac{0.1}{0.12}) \times 10^{6}=e $
$h=0.64 \times 10^{-33}=6.4 \times 10^{-34} J-s$
BETA
