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When a photosensitive surface is illumin...

When a photosensitive surface is illuminated with light of wavelenght `lambda`, the stopping potential 74 is V. When the same surface is illuminated by light of wavelength `2lambda`, stopping potential is V/4. The threshold wavelength for the surface is `N lambda` then N will be ,-

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The correct Answer is:
`3.00`
`e_(v)=(hc)/(lambda)-(hc)/(lambda_(0))`
`(e_(v))/(4)=(hc)/(2lambda)-(hc)/(lambda_(0))`
Dividing
`(4=(1)/(lambda)-(1)/(lambda_(0)))/((1)/(2lambda)-(1)/(lambda_(0)))`
`rArr lambda_(0) = 3lambda`
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