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.