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The fig. shows the variation of photon c...

The fig. shows the variation of photon current with anode potential for a photo-sensitive surface for three different radiation. Let `I_(a), I_(b) and I_(c)` be the intensities and `f_(a), f_(b) and f_(c)` be the frequency for the curves a,b and c respectively.

A

`f_a = f_b and I_a ne I_b`

B

`f_a = f_c and I_a = I_c`

C

`f_a = f_b and I_a = I_b`

D

`f_b = f_c and I_b = I_c`

Text Solution

Verified by Experts

The correct Answer is:
A
Stopping potential depends on frequency and in the given graph , we can see that stopping potential is same for graph a and b , hence we can say that `f_a =f_b` Saturation current for graphs a and b is different and hence intensity for graphs a must be different from that for graph b .
From above discussion , we can understand that option (a) is correct
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The figure shows the variation of photocurrent with anode potential for a photosensitve surface for three different radiations. Let l_a, l_b and l_c be the curves a, b and c, respectively (a) f_a = f_b and l_a != l_b (b) f_a = f_c and l_a = l_c (c ) f_a = f_b and l_a = l_b (d) f_b = f_c and l_b = l_c

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Knowledge Check

  • In the following figure shows the variation of photocurrent with potential of the collector plate for a photosensitive surface for three different radiations. Let I_(a), I_(b) and I_(c) be the intensities and f_(a), f_(b) and f_(c) be the frequencies for the curves a, b, and c, respectively.

    A
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  • Let I_(1) = int_(a)^(b) (int_(a)^(x) f(t) dt) dx and I_(2) =int_(a)^(b) (b-x) f(x) dx then

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