Suppose the wavelength of the incident light is increased from `3000 Å` to `3040Å`. Find the corresponding change in the stopping potential. [Take the product `hc=12.4xx10^(-7)eVm)`]

Answer

`dV_(s)=(hc)/(e ).(dlambda)/(lambda^(2))= -(hc)/(228e)xx10^(7)= -5.5xx10^(-2)"volt"`

Explanation

`V_(s)=(hc)/(elambda)-(W)/(e )`
On differentating `dV_(s)= -(hc)/(elambda^(2))(dlambda)= -5.5xx10^(-2) "volt"`
`dV_(s)=(hc)/(e).(Deltalambda)/(lambda^(2))=(hc)/(228e)xx10^(7)=5.5xx10^(-2)"Volt"`