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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