Question

The PbS electron and hole effective masses are both 0.090 m0 and the dielectric constant is 15.3. Calculate the radius of an exciton in PbS. What is the electron-hole binding energy? How does this energy compare to thermal energy (kT) at room temperature? (Hint: recall that the hydrogen atom has a Bohr radius of 0.053 nm and a binding energy of 13.6 eV)

Answer #1

The exciton Bohr radius of PbS is given by the formula:

- εr = dielectric constant (relative permittivity) (15.3)
- m = mass of electron (0.090mo)
- μ = reduced mass (0.045mo)
- ab = Bohr radius (0.053 nm)

Hence putting all values in the above equation we get,

**a*b= 1.6218 nm**

**The exciton Bohr radius of PbS is 1.6218
nm.**

**Bound Exciton Energy:**

- εr = dielectric constant (relative permittivity) (15.3)
- me = mass of electron (0.090mo)
- μ = reduced mass (0.045mo)
- Ry = Bohr radius (13.6 eV)

Hence putting all values in the above equation we get,

**Eexciton = 0.029 eV**

As we know that **Thermal energy** is approximately
0.025 eV at **room temperature**.

**Hence Eexciton is higher than the thermal energy at
room temperature.**

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