Question

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are E_P, E_Q and E_R, respectively, and they are related by E_P = 2E_Q = 2E_R. 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 ___.

Answer 1

Correct Answer: 6

Step 1: Understanding the Photoelectric Effect Equation

The maximum kinetic energy (E_k) of a photoelectron is given by the equation:

E_k = E_photon - φ

Where:

• E_k is the maximum kinetic energy of the photoelectron.
• E_photon is the energy of the incident photon.
• φ is the work function of the metal (the minimum energy required to release an electron).

Step 2: Relating the Maximum Kinetic Energies of the Photoelectrons

We are given that:

E_P = 2E_Q = 2E_R

So, we can express the kinetic energies of photoelectrons from metals P, Q, and R in terms of one another:

• E_P = 2E_Q
• E_P = 2E_R

Step 3: Work Function and Photon Energy for Metals P, Q, and R

Using the photoelectric effect equation for each metal, we can write the following expressions:

E_P = E_photon,P - φ_P

E_Q = E_photon,Q - φ_Q

E_R = E_photon,R - φ_R

Where:

• φ_P = 4.0 eV (work function of metal P)
• φ_Q = 4.5 eV (work function of metal Q)
• φ_R = 5.5 eV (work function of metal R)

Step 4: Expressing Photon Energies for P, Q, and R

From the relation E_P = 2E_Q, we can substitute the expressions for E_P and E_Q:

E_photon,P - φ_P = 2(E_photon,Q - φ_Q)

Substitute the known values of the work functions (φ_P = 4.0 eV, φ_Q = 4.5 eV):

E_photon,P - 4.0 = 2(E_photon,Q - 4.5)

Simplify:

E_photon,P - 4.0 = 2E_photon,Q - 9.0

E_photon,P = 2E_photon,Q - 5.0

Step 5: Determining the Photon Energy for Metal R

We are also given that E_P = 2E_R. Substituting the expression for E_P:

2E_R = E_photon,P - φ_P

Substitute φ_P = 4.0 eV:

2E_R = E_photon,P - 4.0

Substitute the expression for E_photon,P from earlier:

2E_R = (2E_photon,Q - 5.0) - 4.0

2E_R = 2E_photon,Q - 9.0

Now, use the equation for E_R:

E_R = E_photon,R - φ_R

E_photon,R = E_R + φ_R

Substitute φ_R = 5.5 eV:

E_photon,R = E_R + 5.5

Step 6: Solve for Photon Energy of Metal R

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.