Question

the work function of cesium is 1.96 eV. If radiation of wavelength 4.00 x 10^2 nm is incident on the surface, find the kinetic energy of the ejected photoelectrons in eV and the speed of the ejected electrons

Answer 1

Answer on Question #68379-Physics / Quantum Mechanics

The work function of cesium is $\varphi = 1.96$ eV. If radiation of wavelength $\lambda = 4.00 \times 10^{2}$ nm is incident on the surface, find the kinetic energy of the ejected photoelectrons in eV and the speed of the ejected electrons.

Solution

The maximum kinetic energy of the ejected photoelectrons is given by

$$K_{\max} = \frac{hc}{\lambda} - \varphi.$$

Here $h = 6.62 \times 10^{-34}$ J·s is the Planck constant, $c = 3 \times 10^{8}$ m/s - speed of light.

So

$$\frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4.00 \times 10^{2} \times 10^{-9}} = 4.96 \times 10^{-19} \text{ J} = 3.10 \text{ eV}.$$

The maximum kinetic energy

$$K_{\max} = 3.10 - 1.96 = 1.14 \text{ eV}.$$

The speed of the ejected electrons

$$v = \sqrt{\frac{2K_{\max}}{m}} = \sqrt{\frac{2 \times 1.14 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} = 0.63 \times 10^{6} \frac{\text{m}}{\text{s}}.$$

Answer $K_{\max} = 1.14$ eV, $v = 0.63 \times 10^{6} \frac{\text{m}}{\text{s}}$.

Answer provided by https://www.AssignmentExpert.com