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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ of an emitted electron is given by $E = \frac{hc}{\lambda} - W$,wher

Vedclass · Accepted Answer

$hc = \frac{E_1 - E_2}{\lambda_2 - \lambda_1} \cdot (\lambda_1 \lambda_2)$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ of an emitted electron is given by $E = \frac{hc}{\lambda} - W$,where $W$ is the work function of the metal.For the first case: $E_1 = \frac{hc}{\lambda_1} - W$  --- $(i)$For the second case: $E_2 = \frac{hc}{\lambda_2} - W$  --- (ii)Subtracting equation (ii) from equation $(i)$:$E_1 - E_2 = \left(\frac{hc}{\lambda_1} - W\right) - \left(\frac{hc}{\lambda_2} - W\right)$$E_1 - E_2 = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$$E_1 - E_2 = hc \left(\frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right)$Rearranging to solve for $hc$:$hc = \frac{(E_1 - E_2) \lambda_1 \lambda_2}{\lambda_2 - \lambda_1}$

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When a photosensitive material is illuminated by photons of energy $3.1 \ eV$,the stopping potential of the photoelectrons is $1.7 \ V$. When the same photosensitive material is illuminated by photons of energy $2.5 \ eV$,the stopping potential of the photoelectrons is: (in $V$)

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