Arrange the following cases of photoelectric effect experiments for given wavelength (\( \lambda \)) of incident light and work function (\( \Phi \)) of the metals, in ascending order of the maximum kinetic energies of emitted photoelectrons,
(A) \( \lambda \) = 400 nm \( \Phi \) = 1.9 eV
(B) \( \lambda \) = 250 nm \( \Phi \) = 3.2 eV
(C) \( \lambda \) = 200 nm \( \Phi \) = 4.2 eV
(D) \( \lambda \) = 310 nm \( \Phi \) = 2.4 eV
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (D), (B), (C)
By Einstein's photoelectric equation, the maximum kinetic energy is the photon energy minus the work function.
\( \displaystyle K_{\mathrm{max}} = \frac{hc}{\lambda} - \Phi \)
With \( hc = 1240\,\mathrm{eV\,nm} \), the photon energy in eV is \( 1240/\lambda \) (\( \lambda \) in nm):
(A) \( \lambda \) = 400 nm \( \Phi \) = 1.9 eV — \( K_{\mathrm{max}} = 3.10 - 1.9 = 1.20\,\mathrm{eV} \)
(B) \( \lambda \) = 250 nm \( \Phi \) = 3.2 eV — \( K_{\mathrm{max}} = 4.96 - 3.2 = 1.76\,\mathrm{eV} \)
(C) \( \lambda \) = 200 nm \( \Phi \) = 4.2 eV — \( K_{\mathrm{max}} = 6.20 - 4.2 = 2.00\,\mathrm{eV} \)
(D) \( \lambda \) = 310 nm \( \Phi \) = 2.4 eV — \( K_{\mathrm{max}} = 4.00 - 2.4 = 1.60\,\mathrm{eV} \)
Ascending order: 1.20 < 1.60 < 1.76 < 2.00, that is, (A), (D), (B), (C).