CUET · PHYSICS · PYQ PAPER 2025
The graphs of stopping potential versus frequency of incident radiation for two metals A and B are given.
Identify the correct statement(s) from the following:
(A) The slope of the graphs is equal to the Planck's constant.
(B) Intercept on the (-y) axis is equal to (work function/e).
(C) Threshold frequency for metal A is higher than metal B.
(D) Work function for metal B is greater than metal A.
Choose the correct answer from the options given below:
- A (D) only
- B (A) and (D) only
- C (A) and (C) only
- D (B) and (D) only
Answer & Solution
Correct Answer
(D) (B) and (D) only
Step-by-step Solution
Detailed explanation
\( V_s = \frac{h}{e}f - \frac{\phi}{e} \) Y-intercept \( = -\frac{\phi}{e} \). Intercept on (-y) axis \( = \frac{\phi}{e} \). (B) is correct. From graph, threshold frequency \( f_{0A} \( \phi = hf_0 \). Thus \( \phi_A Correct statements are (B) and (D).
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