When light falls on metal

Here $KE$ is maximum kinetic energy, $h$ is Planck's constant, $\nu$ is frequency and $\nu_{o}$ is the threshold frequency.

$KE_{1}=h\nu_{1}-h\nu_{o}=h(5-\nu_{o}) \\ KE_{2}=h\nu_{2}-h\nu_{o}=h(12-\nu_{o}) \\ \frac{KE_{1}}{KE_{2}}=\frac{5-\nu_{o}}{12-\nu_{o}}=\frac{1}{3} \\ \Rightarrow \nu_{o}=\frac{3}{2}=\boxed{1.5}$

Generalizing:-
Let the frequencies of the lights falling be $f_1$ and $f_2$ respectively such that the kinetic energies of their respective thermions is in the ratio $1:k$ . If $f_n$ is the natural frequency of the given metal and $h$ is planck's constant, then:-

$\begin{aligned} hf_1 & = hf_n + {KE}_1\\ {KE}_1 & = hf_1 - hf_n \longrightarrow \boxed{1}\\ \\ hf_2 & = hf_n + {KE}_2\\ {KE}_2 & = hf_2 - hf_n \longrightarrow \boxed{2}\\ \\ \text{Dividing} \ \boxed{1} \ \text{by} \ \boxed{2}:-\\ \\ \dfrac{{KE}_1}{{KE}_2} & = \dfrac{hf_1 - hf_n}{hf_2 - hf_n}\\ \\ \dfrac{1}{k} & = \dfrac{h\left(f_1 - f_n\right)}{h\left(f_2 - f_n\right)}\\ \\ \dfrac{1}{k} & = \dfrac{f_1 - f_n}{f_2 - f_n}\\ \\ f_2 - f_n & = kf_1 - kf_n\\ \\ kf_n - f_n & = kf_1 - f_2\\ \\ f_n\left(k - 1\right) & = kf_1 - f_2\\ \\ f_n & = \dfrac{kf_1 - f_2}{k - 1} \end{aligned}$

So, we have just got a general formula here! So let us subsitute the values in and obtain our final answer.

We get:-
$\begin{aligned} f_n & = \dfrac{3×5 - 12}{3 - 1}\\ \\ & = \dfrac{15 - 12}{2}\\ \\ & = \dfrac{3}{2}\\ \\ & = \color{#3D99F6}{\boxed{1.50}} \end{aligned}$