A light of wavelength 500 nm falls on a metal surface, whose work function is 2.14 eV, The kinetic energy of the emitted electrons is

0.34 eV

The correct answer is Option (2) → 0.34 eV

Given:

Wavelength, $\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m}$

Work function, $\phi = 2.14 \, \text{eV}$

Energy of incident photon:

$E = \frac{hc}{\lambda}$

$E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{500 \times 10^{-9}}$

$E = 3.978 \times 10^{-19} \, \text{J}$

Convert to eV:

$E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.486 \, \text{eV}$

Kinetic energy of emitted electrons:

$K = E - \phi = 2.486 - 2.14 = 0.346 \, \text{eV}$

Final Answer:

$K = 0.35 \, \text{eV}$