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In photoelectric experiment, if the stopping potential $(V_{0})$ is plotted against $\frac{1}{λ}$ of the incident radiation for two different metals following graph was obtained. Correct statement regarding the graph are

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Work function for metal-I is more than metal-II

When both metals are exposed to radiation of same wavelength, the stopping potential of metal-II is more than metal-I

Slope of the curves for both the metals is same and value is $1.21 \times 10^{- 6} J . m . c^{- 1}$

Work function for metal-II is more than m etal-I

answer is A, C.

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Detailed Solution

$E = W + K . E \Rightarrow \frac{h c}{λ} = \frac{h c}{λ_{0}} + e . V_{0} \to V_{0} = \frac{h c}{e . λ} + (\frac{- h c}{e . λ_{0}})$

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