dual-nature-of-radiation-and-matter Question 16
Question: Q. 9.
Using the graph shown in the figure for stopping potential $V_{s}$ and the incident frequency of photons, calculate Planck’s constant.
A [Delhi I, II, III 2015]
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Solution:
Ans. Kinetic energy of an electron is calculated by,
$$ \text { K.E. }=h v-\phi_{0} $$
Where work function
$$ \phi_{0}=h v_{0} $$
where, $v_{0}$ is cutoff frequency.
From the graph $\quad v_{0}=5 \times 10^{14}$
For frequency $8 \times 10^{14} \mathrm{~Hz}$; K.E. $=e V_{0}$
So, at this frequency,
$$ e V_{0}=h v-h v_{0} $$
$$ \text { So, } \quad h=\frac{e V_{0}}{\Delta v} $$
where, $\Delta v$ is change in frequency
$$ \begin{aligned} & =\frac{1.6 \times 10^{-19} \times 1.23}{3 \times 10^{14}} \ & =0.656 \times 10^{-33} \mathrm{~J}-\mathrm{s} \ & =6.56 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned} $$
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Commonly Made Error
- Several students did not know the correct relations.
- Many candidates were unaware of the fact that intercept of the line give threshold frequency.
