The threshold frequency for a certain photosensitive metal is \( \nu_0 \). When it is illuminated by light of frequency \( \nu=2\nu_0 \), the maximum speed of the photoelectrons is \( v_0 \). What will be the maximum speed of the photoelectrons when the same metal is illuminated by light of frequency \( \nu=5\nu_0 \)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2v_0 \)
By Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons is the photon energy minus the work function \( \phi_0=h\nu_0 \):
\( \displaystyle \frac{1}{2}mv_{\max}^{2}=h\nu-h\nu_0 \)
For light of frequency \( 2\nu_0 \):
\( \begin{aligned} \frac{1}{2}mv_0^{2} &= 2h\nu_0-h\nu_0 \\ &= h\nu_0 \end{aligned} \)
For light of frequency \( 5\nu_0 \):
\( \begin{aligned} \frac{1}{2}mv^{2} &= 5h\nu_0-h\nu_0 \\ &= 4h\nu_0 \end{aligned} \)
Dividing the second result by the first:
\( \displaystyle \frac{v^{2}}{v_0^{2}}=4 \)
\( \Rightarrow\; v=2v_0 \)