MCQ
If frequency of light falling on a photosensitive material doubles
- ASaturation photocurrent doubled
- BSaturation photocurrent becomes more than double
- ✓Cut-off voltage becomes more than double
- DStopping potential doubles
✓
Answer
Correct option: C.
Cut-off voltage becomes more than double
c
(c)
(c)
$h v=E+w_0$
$h v-w_0=E_1$
When frequency is doubled
$2 h v-w_0=E_2$
$\text { or }E_2=2 E_1+w_0$
$E _2$ is more than double of $E_1$
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