1. Light with a wavelength of 500 nm is incident on sodium, which has a work function of 2.28 eV. Determine the maximum kinetic energy of the ejected photoelectron.
Given data:
\[ \nu = \frac{c}{\lambda} \]
\[ \nu = \frac{3.00 \times 10^8}{500 \times 10^{-9}} \]
\[ \nu = 6.00 \times 10^{14} \text{ Hz} \]
\[ E = h\nu \]
\[ E = (6.626 \times 10^{-34} \times 6.00 \times 10^{14}) \]
\[ E = 3.976 \times 10^{-19} \text{ J} \]
Converting to electron volts:
\[ E = \frac{3.976 \times 10^{-19}}{1.602 \times 10^{-19}} \]
\[ E \approx 2.48 \text{ eV} \]
\[ KE_{\text{max}} = E - \phi \]
\[ KE_{\text{max}} = 2.48 - 2.28 \]
\[ KE_{\text{max}} = 0.20 \text{ eV} \]
2. Given a metal with a work function of 6.63 eV, calculate its threshold wavelength.
Given data:
\[ \phi = 6.63 \times 1.602 \times 10^{-19} \]
\[ \phi = 1.062 \times 10^{-18} \text{ J} \]
\[ \lambda_0 = \frac{hc}{\phi} \]
\[ \lambda_0 = \frac{(6.626 \times 10^{-34}) (3.00 \times 10^8)}{1.062 \times 10^{-18}} \]
\[ \lambda_0 = \frac{1.988 \times 10^{-25}}{1.062 \times 10^{-18}} \]
\[ \lambda_0 = 1.87 \times 10^{-7} \text{ m} = 187 \text{ nm} \]
3. For a metal with a threshold wavelength of 12.4 nm, find its work function.
Given data:
\[ \phi = \frac{hc}{\lambda} \]
\[ \phi = \frac{(6.626 \times 10^{-34}) (3.00 \times 10^8)}{12.4 \times 10^{-9}} \]
\[ \phi = \frac{1.988 \times 10^{-25}}{12.4 \times 10^{-9}} \]
\[ \phi = 1.603 \times 10^{-17} \text{ J} \]
\[ \phi = \frac{1.603 \times 10^{-17}}{1.602 \times 10^{-19}} \]
\[ \phi \approx 100 \text{ eV} \]
4. A copper surface is irradiated with light of wavelength 1849 Å, resulting in a stopping potential of 2.7 V. Calculate the threshold frequency and work function for the copper surface.
Given data:
\[ \nu = \frac{c}{\lambda} \]
\[ \nu = \frac{3.00 \times 10^8}{1849 \times 10^{-10}} \]
\[ \nu = 1.62 \times 10^{15} \text{ Hz} \]
\[ \phi = h\nu - eV_s \]
\[ \phi = (6.626 \times 10^{-34} \times 1.62 \times 10^{15}) - (1.602 \times 10^{-19} \times 2.7) \]
\[ \phi = (1.073 \times 10^{-18}) - (4.325 \times 10^{-19}) \]
\[ \phi = 6.41 \times 10^{-19} \text{ J} \]
Converting to electron volts:
\[ \phi = \frac{6.41 \times 10^{-19}}{1.602 \times 10^{-19}} \]
\[ \phi \approx 4.0 \text{ eV} \]
\[ \nu_0 = \frac{\phi}{h} \]
\[ \nu_0 = \frac{6.41 \times 10^{-19}}{6.626 \times 10^{-34}} \]
\[ \nu_0 \approx 9.67 \times 10^{14} \text{ Hz} \]
👉 Try these problems yourself.
Problems on the photoelectric effect test concepts of Einstein’s equation, threshold frequency, stopping potential, and work function. By applying formulas like \( K_{max} = h\nu - \phi \) and \( eV_0 = K_{max} \), students can calculate kinetic energy, photon energy, and emission conditions. These numericals strengthen understanding of quantum theory and its applications in devices such as photoelectric cells and solar panels.