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The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work-function? Justify your answer.

The graph shows variation of stopping potential V_{0} versus frequency of incident radiation *v* for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?

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#### Solution

**eV _{0}=hν−hν_{0}**

`V_0-h/ev-h/ev_0`

Here, *e* is the charge of an electron, *Vo* is the stopping potential, h is Planck's constant, ν_{0} is the threshold frequency and ν is the frequency of the incident light.

On comparing with y=mx+c, we find that the more the magnitude of intercept on *y* or *V _{0}* axis, the more is the threshold frequency (ν

_{0}).

From the graph, the threshold frequency for metal A is greater than that for metal B. Hence, the work function for metal A is greater than that for metal B.

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