Question

If \(\phi\) is the work function of photosensitive material in eV and light of wavelength of numerical value \(\lambda = \frac{hc}{e}\) metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take \(h\)-Planck’s constant, \(c\)-velocity of light in free space) is (in SI units):

• A. \(e + 2\phi\)
• B. \(2e - \phi\)
• C. \(e - \phi\)
• D. \(e + \phi\)

Answer 1

The Correct Option is C

The problem requires us to determine the maximum kinetic energy of a photo-electron ejected when light of a certain wavelength is incident on a photosensitive material. Let's solve this step-by-step using the photoelectric effect formula. 

1. Begin by analyzing the given wavelength \(\lambda\):
   • We are given \(\lambda = \frac{hc}{e}\). This means the energy of the incident light (\(E\)) can be expressed as: \(E = \frac{hc}{\lambda} = e \, \text{Joules}\) (since \(\lambda = \frac{hc}{e}\))
2. We need to identify the work function of the material:
   • The work function \(\phi\) is provided in electron volts (eV), so it has to be converted into joules for calculation: \(\phi \, \text{J} = \phi \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}\).
3. Using the photoelectric effect equation, the kinetic energy (\(K.E.\)) of the ejected electron is given by: \(K.E. = E - \phi\)
4. Substitute the expression for energy from step 1:
   • \(K.E. = e - \phi\)
5. Therefore, the maximum kinetic energy of the ejected photo-electron is \(\boxed{e - \phi}\). This corresponds to the correct answer option given among the choices.

This formula is derived from the conservation of energy, which states that the energy of the incident photon is used to overcome the work function of the material and the rest is converted into the kinetic energy of the electron.

Ruling out other options:

• \(e + 2\phi\): This expression doesn't align with the principle of photoelectric effect as it results in higher energy than available.
• \(2e - \phi\): This improperly doubles the available photon energy, which is incorrect.
• \(e + \phi\): Adds the work function instead of subtracting, which doesn't logically fit the energy balance.