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When photons of energy hv fall on an alu...

When photons of energy `hv` fall on an aluminium plate (of work function `E_(0)`), photoelectrons of maximum kinetic energy `K` are ejected . If the frequency of the radiation is doubled , the maximum kinetic energy of the ejected photoelectrons will be

A

K+hv

B

`K+epsilon_(0)`

C

2K

D

K

Text Solution

Verified by Experts

The correct Answer is:
1
As `K = hv-hv_(0)`
`K_(1)=2hv-hv_(0)=(hv)+(hv)+(hv-hv_(0))`
`k' = hv+K`
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