(a) 263 nm
First of all, let's convert the work function for silver from eV to Joules:
[tex]\phi = 4.73 eV \cdot (1.6\cdot 10^{-19} J/eV)=7.57\cdot 10^{-19} J[/tex]
The energy of the incoming photon is given by:
[tex]E=\frac{hc}{\lambda}[/tex]
where h is the Planck constant, c is the speed of light, [tex]\lambda[/tex] 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,
[tex]E=\phi[/tex]
Substituting and solving for the wavelength,
[tex]\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[/tex]
(b) [tex]1.14\cdot 10^{15}Hz[/tex]
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:
[tex]f = \frac{c}{\lambda}[/tex]
And substituting numbers, we find
[tex]f = \frac{3\cdot 10^8 m/s}{2.63\cdot 10^{-7} m}=1.14\cdot 10^{15}Hz[/tex]
(c) 1.68 eV
The equation for the photoelectric effect is:
[tex]E=\phi + K_{max}[/tex]
where
E is the energy of the incoming photon
[tex]\phi[/tex] is the work function
[tex]K_max[/tex] is the maximum kinetic energy of the photoelectrons
Since
E = 6.41 eV
[tex]\phi = 4.73 eV[/tex]
The maximum kinetic energy of the photoelectrons is
[tex]K_{max}=E-\phi=6.41 eV-4.73 eV=1.68 eV[/tex]
