The graph shows the variation of stopping potential $V_0$ versus frequency of incident radiation $\nu$ for two photosensitive metals A and B. Which of the two metals has the higher threshold frequency and why?

Metal A (its line cuts the $\nu$-axis at the larger frequency).

CBSE Class 12 · 2014 · Modern PhysicsVery Short Answer

The graph shows the variation of stopping potential

V_{0}

versus frequency of incident radiation

ν

for two photosensitive metals A and B. Which of the two metals has the higher threshold frequency and why?

Answer: Metal A (its line cuts the $ν$ -axis at the larger frequency).

The threshold frequency is the intercept of each line on the frequency axis (where

V_{0} = 0

).
The line for metal A meets the axis at a higher frequency (

ν_{0}^{'} > ν_{0}

) than the line for metal B.
A larger threshold frequency means a larger work function, so metal A has the higher threshold frequency.

Metal A (larger frequency-axis intercept)

CBSE Class 12 Board · 2014 · Subjective