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The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be
A
1.2 V
B
2.4 V
C
−1.2 V
D
−2.4 V
Detailed Solution
Energy of the incident photon: $E = \frac{hc}{\lambda} = \frac{1240\ eV\,nm}{200\ nm} = 6.2$ eV
Einstein's equation: $KE_{max} = \frac{hc}{\lambda} - \phi_0 = 6.2 - 5.01 = 1.19$ eV $\approx 1.2$ eV
$eV_0 = KE_{max}$, so the magnitude of the stopping potential is 1.2 V.
To stop the electrons the collector must be made negative with respect to the emitter (a retarding potential), so the applied potential difference is written with a negative sign.
Required potential difference = −1.2 V
Einstein's equation: $KE_{max} = \frac{hc}{\lambda} - \phi_0 = 6.2 - 5.01 = 1.19$ eV $\approx 1.2$ eV
$eV_0 = KE_{max}$, so the magnitude of the stopping potential is 1.2 V.
To stop the electrons the collector must be made negative with respect to the emitter (a retarding potential), so the applied potential difference is written with a negative sign.
Required potential difference = −1.2 V