Answer:
Lithium
Explanation:
The equation for the photoelectric effect is
[tex]\frac{hc}{\lambda}= \phi + K_{max}[/tex]
where
[tex]\frac{hc}{\lambda}[/tex] is the energy of the incident photon, with
h being the Planck constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength of the photon
[tex]\phi[/tex] is the work function of the metal (the minimum energy needed to extract the photoelectron from the metal)
[tex]K_{max}[/tex] is the maximum kinetic energy of the emitted photoelectrons
In this problem, we have
[tex] \lambda= 190 nm = 1.9\cdot 10^{-7}m[/tex] is the wavelength of the incident photon
[tex]K_{max}=4.0 eV[/tex] is the maximum kinetic energy of the electrons
First of all we can find the energy of the incident photon
[tex]E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.90\cdot 10^{-7} m}=1.05\cdot 10^{-18} J[/tex]
Converting into electronvolts,
[tex] E=\frac{1.05\cdot 10^{-18} J}{1.6\cdot 10^{-19} J/eV}=6.6 eV[/tex]
So now we can re-arrange the equation of the photoelectric effect to find the work function of the metal
[tex]\phi = E-K_{max}=6.6 eV - 4.0 eV=2.6 eV[/tex]
So the metal is most likely Lithium, which has a work function of 2.5 eV.
