Question

Light with a wavelength of 600.0 nm is directed at a metallic surface with a work function of 0.96 eV. Find the max kinetic energy and max speed of the photoelectronsWhat is the cutoff potential necessary to stop the photoelectrons? 

Answer 1

According to the Einstein's formula (see https://en.wikipedia.org/wiki/Photoelectric_effect#Theoretical_explanation) the max kinetic energy of the released electrons will be:

$$K = h\nu - W$$

where $W = 0.98eV$ is the work function of the metal, $h = 4.1\times 10^{-15}eV\cdot s$ is the Planck constant, and $\nu$ is the frequency of the light. The frequency is connected wiht wavelength $\lambda = 600nm = 6\times 10^{-7}m$ as follows:

$$\nu = \dfrac{c}{\lambda}$$

where $c = 3\times 10^{8}m/s$ is the speed of light. Subsittuting this into the expression for $K$, obtain:

$$K = \dfrac{hc}{\lambda} - W$$

Substituting the numbers, find:

$$K = \dfrac{4.1\times 10^{-15}\cdot 3\times 10^8}{6\times 10^{-7}} - 0.98 = 1.07eV \approx 1.6\times 10^{-19}J$$

On the other hand, the kinetic energy is given by the following formula (assuming non-relativistic case):

$$K = \dfrac{mv^2}{2}$$

where $m = 9.11\times 10^{-31}kg$ is the mass of electron, and $v$ is its maximum speed. Expressing $v$, obtain:

$$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2\cdot 1.6\times 10^{-19}J}{9.11\times 10^{-31}kg}}\approx 5.93\times 10^5m/s$$

In order to completely reduce this kinetic energy one should apply the following potential difference (according to the energy-work theorem):

$$V = K$$

if $K$ is measured in eV. Then $V$ will be in volts. Thus, obtain:

$$V = 1.07V$$

Answer. Max kinetic energy: $1.07eV$, max speed: $5.93\times 10^5m/s$, stop potential: $1.07V$.