a. What is the color of crystalline CdSe? We know that the bandgap of CdSe is 1.

Question

a. What is the color of crystalline CdSe? We know that the bandgap of CdSe is 1.74 eV. b. What are excitons and exciton Bohr radius? Calculate the exciton Bohr radius in CdSe. c. When are quantum confinement effects observed in nanocrystallites? Explain and calculate the photon energy (in eV) and the wavelength (in nm) in the observation in this picture for L=2 nm?

Eg = 1.74 (eV); m*electron = 0.13; m*hole = 0.45; = 10 for bulk CdSe; = 9 for nano CdSe particles; 0 = 8.85 x 10-12 (F/m); h = 6.626 x 10-34 (J.s); melectron = 9.11 x 10-31 (kg); c = 3 x 108 m/s; e = 1.6 x 10-19 (C); 1 (eV) = 1.6 x 10-19 (J).

Explanation / Answer

a) Eg = 1.74 eV ====> wavelength = 1240/1.74 = 712.64 nm

Thus the color of crystalline CdSe is RED.

b) The optical properties of a material are usually determined by electronic transitions within the material and light scattering effects. Due to Coulomb interaction, the electrons and holes existing in a material are known to form excitons. An exciton is composed of an electron and a hole. The distance between the electron and the hole within an exciton is called Bohr radius of the exciton. Typical exciton Bohr radius of semiconductors is of a few nanometers.

Now reduced mass u = (m*elec)*(m*hole)/[(m*elec) +(m*hole)] =0.13*0.45/0.58 = 0.1 m0

Bohr radius of exciton = a = a0*0*m0/u = 0.053*8.85*10^-12/0.1 = 4.6905 pm

c)  When the length of a semiconductor is reduced to the same order as the exciton radius, i.e., to a few nanometers, quantum confinement effect occurs and the exciton properties are modified. In these confined structures, the exciton nature is modified and novel optical properties are expected.

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