Question

The work function of metal 'A' is greater than that for metal B. The two metals are illuminated with appropriate radiation of frequency $v$ so as to cause photoelectric emission in both metals. If $v_0$ is the threshold frequency and $K_{max}$ is the maximum kinetic energy of photoelectrons, then

• A. $v_0$ for metal $A$ is greater than that for metal $B$
• B. $v_0$ for metal $A$ is less than that for metal $B$
• C. $K_{max}$ for metal $A$ is greater than that for metal $B$
• D. $K_{max}$ for metal $A$ is less than that for metal $B$

Answer 1

Answer: (A), (D)

$v_0$ for metal $A$ is greater than that for metal $B$
$K_{max}$ for metal $A$ is less than that for metal $B$

${\varphi }_A>{\varphi }_B$
$\varphi =h{V}_0$
$V_0{)}_A>V_0{)}_B$
$K_{max}=hV-h{V}_0$
$\therefore K_{max}{)}_A<K_{max}{)}_B$