From the above figure the values of stop...

From the above figure the values of stopping potentials for `M_(1)` and `M_(2)` for a frequency `v_(3)( gt v_(02))` of the incident radiatioins are `V_(1)` and `V_(2)` respectively. Then the slope of the line is equal to

`(V_(2)-V_(1))/(v_(02)-v_(01))`

`(V_(1)-V_(2))/(v_(02)-v_(01))`

`V_(2)/(v_(02)-v_(01))`

`V_(1)/(v_(02)-v_(01))`

Text Solution

Verified by Experts

The correct Answer is:

`hv_(01)+eV_(1)=hv_(02)+eV_(2) e(V_(1)-V_(2))=h(V_(02)-V_(01)),h/e=((V_(1)-V_(2)))/((V_(02)-V_(01))`

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