The work functions of two metals ( $M_{A}$ and $M_{B}$ ) are in the $1:2$ ratio. When these metals a

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The work functions of two metals ( $M_{A}$ and $M_{B}$ ) are in the $1:2$ ratio. When these metals are exposed to photons of energy 6 eV , the kinetic energy of liberated electrons of $M_{A}:M_{B}$ is in the ratio of 2.642 : 1. The work functions (in eV ) of $M_{A}$ and $M_{B}$ are respectively.

Accepted Answer

${KE}_{	ext{max~}} = E - \phi$$$\begin{array}{r} \left( {KE}_{\max} ight)_{1}\ = 6 - \phi_{1}\#(1) \ \left( {KE}_{\max} ight)_{2}\ = 6 - \phi_{2}\#(2) \end{array}$$By eq. (1) divide eq. (2)$${\frac{\left( {KE}_{	ext{max~}} ight)_{1}}{\left( {KE}_{	ext{max~}} ight)_{2}} = \frac{2.642}{1} = \frac{6 - \phi_{1}}{6 - \phi_{2}} }{\frac{2.642}{1} = \frac{6 - \phi_{1}}{6 - 2\phi_{1}} }{\phi_{1} = 2.3eV }$$$\phi_{2} = 4.6eV$.

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