Question

Find the wavelength in (nm) of incident radiation where the work function is 4.12 eV and the stopping potential is 4V. Given that \( h \nu = 1242 \, \text{eV} \cdot \text{nm} \).

Answer 1

Using the photoelectric equation: \[ h \nu = W + e V_{\text{stop}}, \] where: - \( h \nu \) is the energy of the incident photon, - \( W \) is the work function, - \( e \) is the charge of an electron, - \( V_{\text{stop}} \) is the stopping potential. Substitute the given values into the equation: \[ 1242 \, \text{eV} \cdot \text{nm} = 4.12 \, \text{eV} + 4 \, \text{eV}. \] Solve for the energy of the photon: \[ h \nu = 4.12 \, \text{eV} + 4 \, \text{eV} = 8.12 \, \text{eV}. \] Now, use the formula for the energy of a photon: \[ h \nu = \frac{h c}{\lambda}, \] where \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \lambda \) is the wavelength in meters. Substitute the value of \( h \nu \) in eV and convert it to Joules: \[ 8.12 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 1.299 \times 10^{-18} \, \text{J}. \] Now, solve for \( \lambda \): \[ \lambda = \frac{h c}{1.299 \times 10^{-18}}. \] Substitute the known values: \[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3 \times 10^8 \, \text{m/s})}{1.299 \times 10^{-18} \, \text{J}}. \] \[ \lambda \approx 1.53 \times 10^{-7} \, \text{m} = 153 \, \text{nm}. \] Thus, the wavelength of the incident radiation is \( \boxed{153 \, \text{nm}} \).