Light of wavelength 310 nm falls on a metal surface, and the photoelectrons emitted have a maximum kinetic energy of 1.5 eV. The threshold wavelength for the metal is (Take \( hc=1240\,\mathrm{eV\,nm} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 496 nm
By Einstein’s photoelectric equation, \( K_{max}=h\nu-\phi_0 \). Part of the photon energy is used up as the work function. The rest appears as the kinetic energy of the fastest electrons. The threshold wavelength \( \lambda_0 \) is the longest wavelength whose photons can just supply the work function, so \( hc/\lambda_0=\phi_0 \).
The energy of each incident photon is:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{1240\,\mathrm{eV\,nm}}{310\,\mathrm{nm}} \\ &= 4.0\,\mathrm{eV} \end{aligned} \)
The work function of the metal is:
\( \begin{aligned} \phi_0 &= E-K_{max} \\ &= 4.0\,\mathrm{eV}-1.5\,\mathrm{eV} \\ &= 2.5\,\mathrm{eV} \end{aligned} \)
The threshold wavelength follows from the work function:
\( \begin{aligned} \lambda_0 &= \frac{hc}{\phi_0} \\ &= \frac{1240\,\mathrm{eV\,nm}}{2.5\,\mathrm{eV}} \\ &= 496\,\mathrm{nm} \end{aligned} \)