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.

Explanation

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`