Question

$\begin{array}{cc}ColumnI& ColumnII\\ \text{(A) If v (frequency) is increased keeping}\text{I (intensity) and}\Phi \text{(work function) constant}& \text{(P) Stoping potential increases}\\ \left(B\right)\text{If I is increased keeping v and}\Phi \text{constant}& \text{(Q) Saturation photocurrent increases}\\ \text{(C) If the distance between anode and cathode increases.}& \text{(R) Maximum KE of the photoelectrons increase}\\ \text{(D) If}\Phi \text{is decreased keeping v and I constant}& \text{(S) Stopping potential remains the same}\end{array}$

• A. (A) (P,R); (B) (Q,S); (C) (S); (D) (P,R)
• B. (A) (Q,S); (B) (P,R); (C) (S); (D) (P,R)
• C. (A) (P,R); (B) (Q,S); (C) (P,R); (D) (S)
• D. (A) (P,R); (B) (S); (C) (Q,S); (D) (P,R)

Answer 1

The correct option is A

(A) (P,R); (B) (Q,S); (C) (S); (D) (P,R)

From $e{V}_0=hv-\Phi and{K}_{max}=hv-\Phi$. If v increases keeping $\Phi$ constant, then $V_{0}and{K}_{max}$ both increase. If $\Phi$ decreases keeping v constant, then $V_{0}and{K}_{max}$ increase.
If I increases, more photoelectrons would be liberated, hence saturation photocurrent increases.
If the separation between cathode and anode is increased, then there is no effect on $V_0,K_{max}$ or current