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

Using Einstein's photoelectric equation:

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

Given λ = hc/ε.

Thus, ε = hc/λ.

(K.E.)_max = ε - φ