In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are EP, EQ and ER, respectively, and they are related by EP = 2EQ = 2ER. In this experiment, the same source of monochromatic light is used for metals P and Q while a different source of monochromatic light is used for metal R. The work functions for metals P, Q and R are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is ___.
Correct Answer: 6
Solution and Explanation
The maximum kinetic energy (Ek) of a photoelectron is given by the equation:
Ek = Ephoton - φ
Where:
- Ek is the maximum kinetic energy of the photoelectron.
- Ephoton is the energy of the incident photon.
- φ is the work function of the metal (the minimum energy required to release an electron).
We are given that:
EP = 2EQ = 2ER
So, we can express the kinetic energies of photoelectrons from metals P, Q, and R in terms of one another:
- EP = 2EQ
- EP = 2ER
Using the photoelectric effect equation for each metal, we can write the following expressions:
EP = Ephoton,P - φP
EQ = Ephoton,Q - φQ
ER = Ephoton,R - φR
Where:
- φP = 4.0 eV (work function of metal P)
- φQ = 4.5 eV (work function of metal Q)
- φR = 5.5 eV (work function of metal R)
From the relation EP = 2EQ, we can substitute the expressions for EP and EQ:
Ephoton,P - φP = 2(Ephoton,Q - φQ)
Substitute the known values of the work functions (φP = 4.0 eV, φQ = 4.5 eV):
Ephoton,P - 4.0 = 2(Ephoton,Q - 4.5)
Simplify:
Ephoton,P - 4.0 = 2Ephoton,Q - 9.0
Ephoton,P = 2Ephoton,Q - 5.0
We are also given that EP = 2ER. Substituting the expression for EP:
2ER = Ephoton,P - φP
Substitute φP = 4.0 eV:
2ER = Ephoton,P - 4.0
Substitute the expression for Ephoton,P from earlier:
2ER = (2Ephoton,Q - 5.0) - 4.0
2ER = 2Ephoton,Q - 9.0
Now, use the equation for ER:
ER = Ephoton,R - φR
Ephoton,R = ER + φR
Substitute φR = 5.5 eV:
Ephoton,R = ER + 5.5
Since we know the energy must be consistent, the photon energy used for metal R can be found by using the fact that the total energy is conserved:
Therefore, the energy of the incident photon for metal R is 6 eV.
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