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Q.

Radiation of frequency f falls on the surface of two metals A and B. Stopping potential vs frequency f graph for these two metals A and B is shown below
Find ratio $\frac{ϕ_{B} (w o r k f u n c t i o n o f B)}{ϕ_{A} (w o r k f u n c t i o n o f A)}$

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answer is 8.

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Detailed Solution

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${ev}_{0} =hf-w v_{0} = \frac{hf}{e} - \frac{w}{e} ϕ_{B} = 200 h ϕ_{A} = 25 h ∴ \frac{ϕ_{B}}{ϕ_{A}} = 8$

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