Photons of energy 7 ,eV are incident on two m | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = E - \phi$,whe

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$2.0$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = E - \phi$,where $E$ is the energy of the incident photon and $\phi$ is the work function of the metal.For metal $A$: $K_A = 7 \,eV - 6 \,eV = 1 \,eV$.For metal $B$: $K_B = 7 \,eV - 3 \,eV = 4 \,eV$.The de-Broglie wavelength $\lambda$ associated with an electron of kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.Thus,the ratio of wavelengths is $\frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{\sqrt{2mK_A}}}{\frac{h}{\sqrt{2mK_B}}} = \sqrt{\frac{K_B}{K_A}}$.Substituting the values: $\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{4 \,eV}{1 \,eV}} = \sqrt{4} = 2.0$.

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