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 = hν_{0}/e

$ \displaystyle \frac{h\nu_0}{e} = 2 $

hν_{0} = 2eV

φ_{0} = 2eV