Respuesta :
A. Lithium
The equation for the photoelectric effect is:
[tex]E=\phi + K[/tex]
where
[tex]E=\frac{hc}{\lambda}[/tex] is the energy of the incident light, with h being the Planck constant, c being the speed of light, and [tex]\lambda[/tex] being the wavelength
[tex]\phi[/tex] is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)
K is the maximum kinetic energy of the photoelectron
In this problem, we have
[tex]\lambda=190 nm=1.9\cdot 10^{-7}m[/tex], so the energy of the incident light is
[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J[/tex]
Converting in electronvolts,
[tex]E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV[/tex]
Since the electrons are emitted from the surface with a maximum kinetic energy of
K = 4.0 eV
The work function of this metal is
[tex]\phi = E-K=6.5 eV-4.0 eV=2.5 eV[/tex]
So, the metal is Lithium.
B. cesium, potassium, sodium
The wavelength of green light is
[tex]\lambda=510 nm=5.1\cdot 10^{-7} m[/tex]
So its energy is
[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J[/tex]
Converting in electronvolts,
[tex]E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV[/tex]
So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:
Cesium
Potassium
Sodium
C. 4.9 eV
In this case, we have
- Copper work function: [tex]\phi = 4.5 eV[/tex]
- Maximum kinetic energy of the emitted electrons: K = 2.7 eV
So, the energy of the incident light is
[tex]E=\phi+K=4.5 eV+2.7 eV=7.2 eV[/tex]
Then the copper is replaced with sodium, which has work function of
[tex]\phi = 2.3 eV[/tex]
So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be
[tex]K=E-\phi = 7.2 eV-2.3 eV=4.9 eV[/tex]