HELLO,
As per question let assume two cases. In first case frequency is \(\upsilon_1\) and kinetic energy of emitted electron is \(k_1 \). In second case frequency is \(\upsilon_2 \) and kinetic energy of emitted electron is \(k_2 \).
Let work function of the given surface is \(\phi_0\).
Now,
According to energy conversion the kinectic energy of emitted electron is equals to difeerence of energy of a photon and work function.
Mathematically, \(K.E = E - \phi_0\)
and, \(E = h\upsilon\)
Now, \(k_1 = h\upsilon_1-\phi_0\) ------------ i
\(k_2=h\upsilon_2-\phi_0\) ------------ ii
Now, dividing eqn i and ii we get:-
\(\frac{k_1}{k_2} = \frac{h\upsilon_1-\phi_0}{h\upsilon_2-\phi_0}\)
According to question \(\frac{k1}{k2} = \frac{1}{k}\)
and \(\phi_0 = h\upsilon_0\) where \(\upsilon_0 \) is threshold frequency.
Now,
Subsituting the relations we get,
\(\frac{1}{k} = \frac{h(\upsilon_1-\upsilon_0)}{h(\upsilon_2-\upsilon_0)}\)
\(\upsilon_2 - \upsilon_0 = k\upsilon_1-k\upsilon_0\)
\(\upsilon_0(k-1) = k\upsilon_1 - \upsilon_2\)
\(\upsilon_0 = \frac{k\upsilon_1 - \upsilon_2}{k-1}\)
So, the correct option is B.
I HOPE YOU WILL UNDERSTAND.