Question

UV light of 4.13 eV is incident on a photosensitive metal surface having work function 3.13 eV. The maximum kinetic energy of ejected photoelectrons will be

• A. 4.13 eV
• B. 1 eV
• C. 3.13 eV
• D. 7.26 eV

Answer 1

Answer: (B)

1 eV

Answer 2

The energy of the ejected photoelectrons is given by the photoelectric equation:

$K.E. = h\nu - \phi$,

where $h\nu$ is the energy of the incident photons and $\phi$ is the work function of the material.

Given:

$h\nu = 4.13 \, \text{eV}, \quad \phi = 3.13 \, \text{eV}$,

the maximum kinetic energy of the ejected photoelectrons is:

$K.E. = 4.13 \, \text{eV} - 3.13 \, \text{eV} = 1 \, \text{eV}$.

Thus, the correct answer is Option (2).