Question

Ultraviolet light of wavelength 350 nm and intensity \(1.00Wm^{−2 }\) falls on a potassium surface. The maximum kinetic energy of the photoelectron is

• A. 3.3 eV
• B. 1.9 eV
• C. 3.2 eV
• D. 1.3 eV

Hint:

Various data sources may list slightly different work function values for potassium, from about 2.2eV up to 2.3eV. Here, choosing ϕ ≈ 2.25 eV leads to a final answer of 1.30eV, matching option (d).

Answer 1

The Correct Option is D

Energy of incident photon, $E_\gamma = \frac{hc}{\lambda}$.
Using $h = 6.626 \times 10^{-34} \text{ J s}, c = 3.0 \times 10^8 \text{ m/s}, \lambda = 350 \times 10^{-9} \text{ m},$ and $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$, we get:

$E_\gamma = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m/s})}{350 \times 10^{-9} \text{ m}} $

 $\approx 5.68 \times 10^{-19} \text{ J} $

$\approx \frac{5.68 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} $ 

$\approx 3.55 \text{ eV}. $
If the work function $(\phi)$ of potassium is taken to be $\approx 2.25 \text{ eV}$ (a commonly cited value in some references), then the maximum kinetic energy of the photoelectrons is
 $K_{\text{max}} = E_\gamma - \phi  $
 $= 3.55 \text{ eV} - 2.25 \text{ eV} $
 $= 1.30 \text{ eV}. $
Thus the maximum kinetic energy is approximately $\boxed{1.3 \text{ eV}}$.