When a piece of metal is illuminated by monochromatic light of wavelength '\( \lambda \)', the stopping potential for photoelectric current is \( 2.5 V_0 \). When the same surface is illuminated by light of wavelength \( 1.5 \lambda \), the stopping potential becomes '\( V_0 \)'. The value of threshold wavelength for photoelectric emission is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2.25 \lambda \)
Einstein’s photoelectric equation gives \( eV_s = \frac{hc}{\lambda} - \phi_0 \), where \( \phi_0 = hc/\lambda_0 \) is the work function of the metal and \( \lambda_0 \) the threshold wavelength. Write it for the two cases.
\( \displaystyle e(2.5V_0) = \frac{hc}{\lambda} - \phi_0 \qquad (1) \)
\( \displaystyle eV_0 = \frac{hc}{1.5\lambda} - \phi_0 = \frac{2hc}{3\lambda} - \phi_0 \qquad (2) \)
Subtracting (2) from (1) removes \( \phi_0 \).
\( \displaystyle 1.5\,eV_0 = \frac{hc}{\lambda}\left(1 - \frac{2}{3}\right) = \frac{hc}{3\lambda} \Rightarrow eV_0 = \frac{2hc}{9\lambda} \)
Putting this in (2) gives the work function.
\( \displaystyle \phi_0 = \frac{2hc}{3\lambda} - \frac{2hc}{9\lambda} = \frac{4hc}{9\lambda} \)
Since \( \phi_0 = hc/\lambda_0 \):
\( \displaystyle \lambda_0 = \frac{9}{4}\lambda = 2.25\lambda \)