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 maximum kinetic energy (K.E.) of emitted electrons, as per Einstein's photoelectric equation, is calculated as:

(K.E.)_max = ε - φ, where ε represents the energy of the incident photon (in eV) and φ denotes the work function.

The energy of the incident photon (ε) can be expressed as:

ε = hc/λ, where λ is the wavelength of the incident light.

Substituting this into the photoelectric equation gives:

(K.E.)_max = ε - φ