Ultraviolet light of wavelength 350 nm and intensity \(1.00Wm^{−2 }\) falls on a potassium surface. The maximum kinetic energy of the photoelectron is
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 $\boxed{1.3 \text{ eV}}$.
List I | List II |
---|---|
(A) (∂S/∂P)T | (I) (∂P/∂T)V |
(B) (∂T/∂V)S | (II) (∂V/∂S)P |
(C) (∂T/∂P)S | (III) -(∂V/∂T)P |
(D) (∂S/∂V)T | (IV) -(∂P/∂S)V |