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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - hf_0$,where $f$ is the frequency o

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$3$

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = hf - hf_0$,where $f$ is the frequency of incident radiation and $f_0$ is the threshold frequency.For the first experiment: $K_1 = h(f_1 - f_0) = h(4 	imes 10^{15} - f_0)$.For the second experiment: $K_2 = h(f_2 - f_0) = h(6 	imes 10^{15} - f_0)$.Given the ratio $K_1 : K_2 = 1 : 3$,we have $3K_1 = K_2$.Substituting the expressions: $3h(4 	imes 10^{15} - f_0) = h(6 	imes 10^{15} - f_0)$.Dividing by $h$: $3(4 	imes 10^{15} - f_0) = 6 	imes 10^{15} - f_0$.$12 	imes 10^{15} - 3f_0 = 6 	imes 10^{15} - f_0$.$12 	imes 10^{15} - 6 	imes 10^{15} = 3f_0 - f_0$.$6 	imes 10^{15} = 2f_0$.$f_0 = 3 	imes 10^{15} \,Hz$.Thus,the threshold frequency is $3 	imes 10^{15} \,Hz$.

When the electromagnetic radiations of frequencies $4 \times 10^{15} \,Hz$ and $6 \times 10^{15} \,Hz$ fall on the same metal in different experiments,the ratio of maximum kinetic energy of electrons liberated is $1: 3$. The threshold frequency for the metal is ............... $\times 10^{15} Hz$.

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