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Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is:
A
–1 V
B
–3 V
C
+3 V
D
+4 V
Explanation
Work function 3 eV; 6 eV photons need 3 V retarding potential.
Detailed Solution
$eV_s = \frac{1}{2}mv_{max}^2 = h\nu - \phi_0$
$2 = 5 - \phi_0 \Rightarrow \phi_0 = 3$ eV
In the second case: $eV_s = 6 - 3 = 3$ eV $\Rightarrow V_s = 3$ V
$\therefore V_{AC} = -3$ V
$2 = 5 - \phi_0 \Rightarrow \phi_0 = 3$ eV
In the second case: $eV_s = 6 - 3 = 3$ eV $\Rightarrow V_s = 3$ V
$\therefore V_{AC} = -3$ V