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

Answer: (C)

$e - \phi$

Answer 2

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