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A metal surface having a work function p...

A metal surface having a work function `phi = 2.2 xx 10^(-19) J`, is illuminated by the light of wavelengh `1320Å`. What is the maximum kinetic energy ( in eV ) of the emitted photoelectron ? [Take `h = 6.6 xx 10^(-34) Js `]

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To solve the problem, we will follow these steps: ### Step 1: Understand the relationship between energy, work function, and kinetic energy The maximum kinetic energy (K.E. max) of the emitted photoelectron can be calculated using the equation derived from the photoelectric effect: \[ K.E._{\text{max}} = E - \phi \] where: - \(E\) is the energy of the incoming photons, - \(\phi\) is the work function of the metal. ### Step 2: Calculate the energy of the incoming photons The energy \(E\) of the photons can be calculated using the formula: \[ E = \frac{hc}{\lambda} \] where: - \(h\) is Planck's constant, - \(c\) is the speed of light, - \(\lambda\) is the wavelength of the light. Given: - \(h = 6.6 \times 10^{-34} \, \text{Js}\), - \(c = 3 \times 10^8 \, \text{m/s}\), - \(\lambda = 1320 \, \text{Å} = 1320 \times 10^{-10} \, \text{m}\). ### Step 3: Substitute the values to find \(E\) Substituting the values into the energy formula: \[ E = \frac{(6.6 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{1320 \times 10^{-10} \, \text{m}} \] Calculating \(E\): \[ E = \frac{(6.6 \times 3) \times 10^{-34 + 8 + 10}}{1320} \] \[ E = \frac{19.8 \times 10^{-16}}{1320} \approx 1.5 \times 10^{-18} \, \text{J} \] ### Step 4: Calculate the maximum kinetic energy Now, we substitute \(E\) and \(\phi\) into the kinetic energy formula: \[ K.E._{\text{max}} = E - \phi \] Given \(\phi = 2.2 \times 10^{-19} \, \text{J}\): \[ K.E._{\text{max}} = (1.5 \times 10^{-18} \, \text{J}) - (2.2 \times 10^{-19} \, \text{J}) \] \[ K.E._{\text{max}} = 1.5 \times 10^{-18} - 0.22 \times 10^{-18} = 1.28 \times 10^{-18} \, \text{J} \] ### Step 5: Convert kinetic energy from Joules to electron volts To convert Joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\): \[ K.E._{\text{max}} \, (\text{in eV}) = \frac{1.28 \times 10^{-18} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 8 \, \text{eV} \] ### Final Answer The maximum kinetic energy of the emitted photoelectron is approximately \(8 \, \text{eV}\). ---
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