The threshold frequency of a photoelectric metal is \( \nu_0 \). If light of frequency \( 2\nu_0 \) is incident on the surface of that metal, maximum velocity of emitted electrons is \( v \). What will be the maximum velocity of emitted electrons if light of frequency \( 4\nu_0 \) is incident on the same metal surface.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \sqrt{3}\,v \)
By Einstein's photoelectric equation, the maximum kinetic energy is the photon energy minus the work function \( h\nu_0 \).
\( \displaystyle \frac{1}{2}mv^{2} = h(2\nu_0) - h\nu_0 = h\nu_0 \)
\( \displaystyle \frac{1}{2}mv'^{2} = h(4\nu_0) - h\nu_0 = 3h\nu_0 \)
Dividing the second equation by the first:
\( \displaystyle \frac{v'^{2}}{v^{2}} = 3 \;\Rightarrow\; v' = \sqrt{3}\,v \)