Q: From the graph shown, the value of Work function if the stopping potential (V), and frequency of the incident light, v, are on y and x-axes respectively is
(a) 1 eV
(b) 2 eV
(c) 4 eV
(d) 3 eV
Ans:(b)
Sol: $ \displaystyle h\nu = \phi + K_{max} $
$ \displaystyle h\nu = h\nu_0 + eV $
$ \displaystyle V = \frac{h\nu}{e} – \frac{h\nu_0}{e} $
On comparing with y = mx + c
Slope = h/e , Intercept = hν0/e
$ \displaystyle \frac{h\nu_0}{e} = 2 $
hν0 = 2eV
φ0 = 2eV