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When monochromatic light of wavelength `lambda` is incident on a metallic surface the stopping potential for photoelectric current is `3V_(0)` when same surface is illuminated with light of waelength `2lambda` the stopping potential is `V_(0)`
The threshold wavelength for this surface when photoelectric effect takes place is

A

`lambda`

B

`2 lambda`

C

`3 lambda`

D

`4 lambda`

Text Solution

Verified by Experts

The correct Answer is:
d

From photoelectric equation
`hv-hv_(0)=ev_(0) ro (hc)/(lambda)-(hc)/(lambda_(0))=ev_(0)`
on dividing eq (i) by eq (ii) we get
`(lambda_(0)-lambda)/(lambda_(0)-2lambda)//2=3/1`
or `2lambda_(0)-2lambda=3lambda_(0)-6lambda`
`rarr lambda_(0)= 4 lambda`
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