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(B) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = h
u - W_0$.For metal $A$: $K_A = hf - W_A$.For

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$1:2$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{\max} = h
u - W_0$.For metal $A$: $K_A = hf - W_A$.For metal $B$: $K_B = h(2f) - W_B$.Given the ratio of work functions $\frac{W_A}{W_B} = \frac{1}{2}$,let $W_A = W$ and $W_B = 2W$.Assuming the threshold frequency is such that $hf > W_A$ and $h(2f) > W_B$,we look for the ratio $\frac{K_A}{K_B}$.However,without specific values for $hf$ relative to $W$,the ratio depends on the incident energy. If we assume $hf = 2W$ (a common convention for such problems where the ratio is independent of $f$),then:$K_A = 2W - W = W$.$K_B = h(2f) - 2W = 2(2W) - 2W = 4W - 2W = 2W$.Therefore,the ratio $\frac{K_A}{K_B} = \frac{W}{2W} = 1:2$.

The ratio of work functions of two metals is $1:2$. If light of frequencies $f$ and $2f$ are incident on them respectively,what is the ratio of the maximum kinetic energy of the emitted photoelectrons?

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