Question

# The PbS electron and hole effective masses are both 0.090 m0 and the dielectric constant is...

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)

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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