The work functions of metals A and B are in t | vedclass

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(B) Given the ratio of work functions: $\frac{\phi_{A}}{\phi_{B}} = \frac{1}{2}$,which implies $\phi_{B} = 2\phi_{A}$.According to Einstein's photoele

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(B) Given the ratio of work functions: $\frac{\phi_{A}}{\phi_{B}} = \frac{1}{2}$,which implies $\phi_{B} = 2\phi_{A}$.According to Einstein's photoelectric equation,the maximum kinetic energy $E_{K}$ is given by $E_{K} = h
u - \phi$.For metal $A$: $E_{K1} = hf - \phi_{A}$  --- $(1)$For metal $B$: $E_{K2} = h(2f) - \phi_{B} = 2hf - 2\phi_{A}$ --- $(2)$From equation $(1)$,we have $hf = E_{K1} + \phi_{A}$.Substituting this into equation $(2)$:$E_{K2} = 2(E_{K1} + \phi_{A}) - 2\phi_{A}$$E_{K2} = 2E_{K1} + 2\phi_{A} - 2\phi_{A}$$E_{K2} = 2E_{K1}$Therefore,the ratio $\frac{E_{K1}}{E_{K2}} = \frac{1}{2}$.

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