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When photons of energy hν fall on an aluminium plate (of work function E0), photo electrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photo electrons will be
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- 2K
- K
- K + hν
- K + E0
Correct Option: C
Applying Einstein's formula for photo-electricity
hv = φ + | mv2 ; | 2 |
hv = φ + K
φ = hv - K
If we use 2ν frequency then let the kinetic energy becomes K'
So,h . 2ν = φ + K'
2hν = hν – K + K'
K' = hν + K