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Figure is the plot of the 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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Answer» Solution :We have to take two cases Case - I`v_0 = 1.656 ``v = 5 XX 10^(14) Hz` Case -II ` v_0 = 0 ` ` v = 1 xx 10^(14) Hz` (b) We know ` ev_0 = hv - W_0 ` ` 1.656 e =H xx 5 xx 10^(14) - W_0 ` `0 = 5h xx 10^(14) - 5W_0 ` ` 1.6556 e = 4W_0` ` rArr W_0 = 1.656 / 4 eV ` ` = 0.414eV ` (a) Putting the VALUE of W_0 in equation (2) ` rArr 5 W_0 = 5h xx 10^(14) ` `rArr 5 xx 0.414= 5 xx h xx 10^(14)` ` rArr h = 4.414 xx 10^(-15) eV-s` |
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