Answer:
[tex]5.04\cdot 10^8 A[/tex]
Explanation:
The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:
[tex]\phi = 3.950\cdot 10^{-19}J[/tex]
So, the energy of the incoming photon hitting on the metal must be at least equal to this value.
The energy of a photon is given by
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck's constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength of the photon
Using [tex]E=\phi[/tex] and solving for [tex]\lambda[/tex], we find the maximum wavelength of the radiation that will eject electrons from the metal:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m[/tex]
And since
1 angstrom = [tex]10^{-15}m[/tex]
The wavelength in angstroms is
[tex]\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A[/tex]