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Question
The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.
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Solution
We have to take two cases.
Case (I)
When stopping potential, `V_0 = 1.656 "Volts"`
Frequency , `v = 5 xx 10^14 "Hz"`
Case (II) When stopping potential , `V_0 = 0`
Frequency , `v=1 xx 10^14 "Hz"`
(b)From Einstein's equation,
`eV_0= hv- W_0`
On substituting the values of case(1) and case(2), we get:
`1.656e= h xx 5 xx 10^14- W_0` ...(1)
`0 = 5 xx h xx 1 xx 10^14 - 5 xx W_0` ...(2)
Subtracting equation(2) from (1), we get:
`W_0= 1.656/4 eV`
= 0.414 eV
(a) Putting the value of W0 in equation (2), we get:
`5W_0= 5h xx 10^14`
`5 xx 0.414= 5 xx h xx 10^14`
`h= 4.414 xx 10^-15 "eVs"`
Or
`h/e= 4.414 xx 10^-15 "Vs"`
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