Question

The work function for silver is 4.73 eV. (a) Find the cutoff wavelength for silver. (b) What is the lowest frequency of light incident on silver that releases photoelectrons from its surface? Hz (c) If photons of energy 6.41 eV are incident on silver, what is the maximum kinetic energy of the ejected photoelectrons? eV

Answer 1

(a) 263 nm

First of all, let's convert the work function for silver from eV to Joules:

$\phi = 4.73 eV \cdot (1.6\cdot 10^{-19} J/eV)=7.57\cdot 10^{-19} J$

The energy of the incoming photon is given by:

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

where h is the Planck constant, c is the speed of light, $\lambda$ is the photon's wavelength.

The cutoff wavelength is the minimum wavelength for which the photon has enough energy to extract the photoelectron from the material: that means, the wavelength at which the energy of the photon is at least equal to the work function of the material,

$E=\phi$

Substituting and solving for the wavelength,

$\frac{hc}{\lambda}=\phi\\\lambda=\frac{hc}{\phi}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{7.57\cdot 10^{-19} J}=2.63\cdot 10^{-7} m = 263 nm$

(b) $1.14\cdot 10^{15}Hz$

The lowest frequency of light incident on silver that releases photoelectrons from its surface is the frequency corresponding to the wavelength we found at point (a); using the relationship between frequency and wavelength:

$f = \frac{c}{\lambda}$

And substituting numbers, we find

$f = \frac{3\cdot 10^8 m/s}{2.63\cdot 10^{-7} m}=1.14\cdot 10^{15}Hz$

(c) 1.68 eV

The equation for the photoelectric effect is:

$E=\phi + K_{max}$

where

E is the energy of the incoming photon

$\phi$ is the work function

$K_max$ is the maximum kinetic energy of the photoelectrons

Since

E = 6.41 eV

$\phi = 4.73 eV$

The maximum kinetic energy of the photoelectrons is

$K_{max}=E-\phi=6.41 eV-4.73 eV=1.68 eV$