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When a surface 1 cm thick is illuminated...

When a surface 1 cm thick is illuminated with light of wavelength `lambda`, the stopping potential is `V_(0)`, but when the same surface is illuminated by light of wavelength `3lambda`, the stopping potential is `(V_(0))/(6)`. Find the threshold wavelength for metallic surface.

A

`4lambda`

B

`5lambda`

C

`3lambda`

D

`2lambda`

Text Solution

Verified by Experts

The correct Answer is:
B
`eV_(0)=hc[(1)/(lambda)-(1)/(lambda_(0))]" "....(i)`
`(eV_(0))/(6)=hc[(1)/(3lambda)-(1)/(lambda_(0))]" "...(ii)`
Dividing Eq. (i) Eq. (ii), we get
`6=(((1)/(lambda)-(1)/(lambda_(0))))/(((1)/(3lambda)-(1)/(lambda_(0))))`
`(6)/(3lambda)-(6)/(lambda_(0))=(1)/(lambda)-(1)/(lambda_(0))`
`(1)/(lambda)=(5)/(lambda_(0))`
`lambda=(lambda_(0))/(5)`
`"or "lambda_(0)=5lambda`
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