Question

The graph between the stopping potential $V_0$ and frequency $v$ for two different metal plates P and Q are shown in the figure. Which of the metals has greater threshold wavelength and work function?

Answer 1

The threshold frequency of metal plate P is lesser than that of Q.

Since $\lambda \propto \frac{1}{\nu }$, the threshold wavelength of metal plate P is greater than that of Q.

The work function of a metal is given as

$W_0=h{\nu }_0$

Since the threshold frequency of metal Q is greater, the work function it is also greater than that of P.