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If the frequency of light falling on a metal plate is doubled, the kinetic energy of emitted electrons will be:
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- Exactly double
- Sightly more than double
- Slightly less than double
- Four times
Correct Option: C
Kinetic energy is given by
| K. E. = | mv2 = hv – hν0 = hν – φ0 | |
| 2 |
where, hν0 = φ0 = work function i.e. minimum photonic energy to liberate electron from metallic bonding.
ν = falling frequency
| Thus, | = | ||
| (K.E.)1 | hν – φ0 |
from above relation, we can say that if the frequency of light falling on a metal plate is doubled, the kinetic energy of the emitted electrons will be slightly less than double.
