A. Lithium
The equation for the photoelectric effect is:
![E=\phi + K](https://tex.z-dn.net/?f=E%3D%5Cphi%20%2B%20K)
where
is the energy of the incident light, with h being the Planck constant, c being the speed of light, and
being the wavelength
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
, so the energy of the incident light is
![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](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D%3D%5Cfrac%7B%286.63%5Ccdot%2010%5E%7B-34%7DJs%29%283%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B1.9%5Ccdot%2010%5E%7B-7%7D%20m%7D%3D1.05%5Ccdot%2010%5E%7B-18%7DJ)
Converting in electronvolts,
![E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1.05%5Ccdot%2010%5E%7B-18%7DJ%7D%7B1.6%5Ccdot%2010%5E%7B-19%7D%20J%2FeV%7D%3D6.5%20eV)
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
![\phi = E-K=6.5 eV-4.0 eV=2.5 eV](https://tex.z-dn.net/?f=%5Cphi%20%3D%20E-K%3D6.5%20eV-4.0%20eV%3D2.5%20eV)
So, the metal is Lithium.
B. cesium, potassium, sodium
The wavelength of green light is
![\lambda=510 nm=5.1\cdot 10^{-7} m](https://tex.z-dn.net/?f=%5Clambda%3D510%20nm%3D5.1%5Ccdot%2010%5E%7B-7%7D%20m)
So its energy is
![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](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D%3D%5Cfrac%7B%286.63%5Ccdot%2010%5E%7B-34%7DJs%29%283%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B5.1%5Ccdot%2010%5E%7B-7%7D%20m%7D%3D3.9%5Ccdot%2010%5E%7B-19%7DJ)
Converting in electronvolts,
![E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B3.9%5Ccdot%2010%5E%7B-19%7DJ%7D%7B1.6%5Ccdot%2010%5E%7B-19%7D%20J%2FeV%7D%3D2.4%20eV)
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: ![\phi = 4.5 eV](https://tex.z-dn.net/?f=%5Cphi%20%3D%204.5%20eV)
- Maximum kinetic energy of the emitted electrons: K = 2.7 eV
So, the energy of the incident light is
![E=\phi+K=4.5 eV+2.7 eV=7.2 eV](https://tex.z-dn.net/?f=E%3D%5Cphi%2BK%3D4.5%20eV%2B2.7%20eV%3D7.2%20eV)
Then the copper is replaced with sodium, which has work function of
![\phi = 2.3 eV](https://tex.z-dn.net/?f=%5Cphi%20%3D%202.3%20eV)
So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be
![K=E-\phi = 7.2 eV-2.3 eV=4.9 eV](https://tex.z-dn.net/?f=K%3DE-%5Cphi%20%3D%207.2%20eV-2.3%20eV%3D4.9%20eV)