1.

When a metallic surface is illuminated with radiation of wavelength lambda,the stopping potential is V.If the same surface is illuminated with radiation of wavelength 2lambda,the metallic surface is

Answer»

5 `LAMBDA`
`(5)/(2)lambda`
`3lambda`
`4LAMBDA`

Solution :For first case,
`EV=(hc)/(lambda)-(hc)/(lambda_(0))`
For second case
`e(V)/(4)=(hc)/(2lambda)-(hc)/(lambda_(0))`
`therefore eV=(2hc)/(lambda)-(4hc)/(lambda_(0))` ……..(2)
By comparing (1) and (2)
`therefore (hc)/(lambda)-(hc)/(lambda_(0))=(2hc)/(lambda)-(4hc)/(lambda_(0))`
`therefore (hc)/(lambda)-(2hc)/(lambda)=(hc)/(lambda_(0))-(4hc)/(lambda_(0))`
`therefore -(hc)/(lambda)=-(3hc)/(lambda_(0))`
`therefore -(hc)/(lambda)=-(3hc)/(lambda_(0))`
`therefore lambda_(0)=3lambda`


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