Arrange the cut-off wavelengths for the following photosensitive metals in ascending order. Their work functions are given inside the bracket:
(Take: Planck’s constant \( = 6.6\times10^{-34} \) J s; speed of light \( = 3\times10^{8}\,\mathrm{m\,s^{-1}} \))
(A) Silver (\( 4.7 \) eV)
(B) Sodium (\( 2.3 \) eV)
(C) Copper (\( 4.5 \) eV)
(D) Cesium (\( 1.9 \) eV)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C), (B), (D)
The cut-off (threshold) wavelength is the longest wavelength that can eject photoelectrons: \( hc/\lambda_{0} = \phi_{0} \), so \( \lambda_{0} = hc/\phi_{0} \). A smaller work function gives a longer cut-off wavelength.
\( \displaystyle hc = \frac{(6.6\times10^{-34})(3\times10^{8})}{1.6\times10^{-19}}\,\mathrm{eV\,m} \approx 1.24\times10^{-6}\,\mathrm{eV\,m} = 1240\,\mathrm{eV\,nm} \)
\( \text{(A) Silver: } \lambda_{0} = 1240/4.7 \approx 264\,\mathrm{nm} \)
\( \text{(C) Copper: } \lambda_{0} = 1240/4.5 \approx 276\,\mathrm{nm} \)
\( \text{(B) Sodium: } \lambda_{0} = 1240/2.3 \approx 539\,\mathrm{nm} \)
\( \text{(D) Cesium: } \lambda_{0} = 1240/1.9 \approx 653\,\mathrm{nm} \)
Ascending order of cut-off wavelength: (A), (C), (B), (D).
Shortcut: Since \( \lambda_{0} \propto 1/\phi_{0} \), no calculation is needed: arrange the work functions in descending order (4.7, 4.5, 2.3, 1.9 eV).