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The surface of a metal is illuminated alternatively with photons of energies E=4eV and E=2.5eV resp. The ratio of maximum speeds of the photoelectrons emitted is the two cases is 2. The work function of  the metal in eV is ......
Option: 1 2.5
Option: 2 5
Option: 3 7.5
Option: 4 10

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Correct Answer: 1

\mathrm{E}_{1}=\phi+\mathrm{K}_{1} \ldots(1)$ \\ $\mathrm{E}_{2}=\phi+\mathrm{K}_{2}$ $\ldots(2)$ \\ $E_{1}-E_{2}=K_{1}-K_{2}$ \\ Now $\frac{V_{1}}{V_{2}}=2$ \\ $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=4$ \\ $\mathrm{K}_{1}=4 \mathrm{~K}_{2}$ \\ Now from equation (2) $\Rightarrow 4-2.5=4 \mathrm{~K}_{2}-\mathrm{K}_{2}$ \\ $1.5=3 \mathrm{~K}_{2}$ \\\ $\mathrm{K}_{2}=0.5 \mathrm{eV}$ \\ Now putting This Value in equation (2) \\ $2.5=\phi+0.5 \mathrm{eV}$ \\ $\phi=2 \mathrm{ev}$

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Deependra Verma

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