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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$

Vedclass · Accepted Answer

$\frac{\lambda_{1} E_{1}-\lambda_{2} E_{2}}{\lambda_{2}-\lambda_{1}}$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For wavelength $\lambda_{1}$,$E_{1} = \frac{hc}{\lambda_{1}} - \phi$  --- $(i)$For wavelength $\lambda_{2}$,$E_{2} = \frac{hc}{\lambda_{2}} - \phi$  --- (ii)From $(i)$,$hc = \lambda_{1}(E_{1} + \phi)$.From (ii),$hc = \lambda_{2}(E_{2} + \phi)$.Equating the two expressions for $hc$:$\lambda_{1}(E_{1} + \phi) = \lambda_{2}(E_{2} + \phi)$$\lambda_{1}E_{1} + \lambda_{1}\phi = \lambda_{2}E_{2} + \lambda_{2}\phi$$\phi(\lambda_{1} - \lambda_{2}) = \lambda_{2}E_{2} - \lambda_{1}E_{1}$$\phi = \frac{\lambda_{2}E_{2} - \lambda_{1}E_{1}}{\lambda_{1} - \lambda_{2}}$Multiplying numerator and denominator by $-1$ gives $\phi = \frac{\lambda_{1}E_{1} - \lambda_{2}E_{2}}{\lambda_{2} - \lambda_{1}}$.

When a photosensitive surface is irradiated by light of wavelengths $\lambda_{1}$ and $\lambda_{2}$,the kinetic energies of the emitted photoelectrons are $E_{1}$ and $E_{2}$ respectively. The work function of the photosensitive surface is

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