Question

The figure shows stopping potential $V_0$ and frequency $\upsilon$ for two different metallic surfaces A and B. The work function of A, as compared to that of B is:

• A. less
• B. more
• C. equal
• D. nothing can be said

Answer 1

Answer: (A)

less

$eV=h\upsilon -\varphi$

From above equation,

The value of threshold frequency ${\nu }_0$ for A is less than that for B , hence ${\varphi }_A$ < ${\varphi }_B$.