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]

Show Answer

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

$$ \begin{aligned} & =\frac{1.6 \times 10^{-19} \times 1.23}{3 \times 10^{14}} \ & =0.656 \times 10^{-33} \mathrm{~J}-\mathrm{s} \ & =6.56 \times 10^{-34} \mathrm{~J}-\mathrm{s} \end{aligned} $$

1

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.


विषयसूची

जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक