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The threshold frequency for a certain ph...

The threshold frequency for a certain photosensitive metal is `v_(0)`. When it is illuminated by light of frequency `v=2 v_(0)`, the maximum velocity of photoelectrons is `v_(0)`.The maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency `v=5 v_(0)` is `x v_(0)`. Find `x`

A

`sqrt2v_0`

B

`2v_0`

C

`2sqrt2v_0`

D

`4v_0`

Text Solution

Verified by Experts

The correct Answer is:
B
`k_("max") = hv - phi_0 `
`1/2 mv_0^2 = h (2v_0) - hv_0 = hv_0" "...(i)`
`1/2 mv_1^2 = h (5v_0) - hv_0 = 4hv_0" "...(ii)`
From (i) and (ii) –
`v_1^2/v_0^2=4/1 impliesv_1^2=4v_0^2impliesv_1 =2v_0`
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