Figure shows the variation of stopping potential with frequency (?) for the two photosensitive materials and . The slope gives the value of , where h is Planck’s constant, e is the charge of an electron.

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Solution

Q.
Given below are two statements:

Statement I: Figure shows the variation of stopping potential with frequency (ν) for the two photosensitive materials $M_{1}$ and $M_{2}$ . The slope gives value of $\frac{h}{e}$ , where h is Planck’s constant, e is the charge of an electron.

Statement II: $M_{2}$ will emit photoelectrons of greater kinetic energy for the incident radiation having the same frequency.

In the light of the above statements, choose the most appropriate answer from the options given below. [2024]

1 Statement I is incorrect but Statement II is correct

2 Both Statement I and Statement II are incorrect

3 Both Statement I and Statement II are correct

4 Statement I is correct and Statement II is incorrect

Ans.
(4)

$e V_{s} = h ν + ϕ \Rightarrow V_{s} = \frac{h}{e} ν + \frac{ϕ}{e}$

slope = $\frac{h}{e}$

Work function of $M_{2}$ is higher than $M_{1}$ so kinetic energy of emitted electron is less for $M_{2}$

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