Question

Consider an n-doped semiconductor with an effective electron mass equal to one half the free electron mass, and a dielectric constant of 2. Which of the following statements is correct? More than one is correct

A. The impurity will look like a hydrogen atom, but with a hole circling it instead of an electron
B. The impurities will look like Hydrogen atoms, but smaller in size than a Hydrogen atom in free space
C. The impurity will look like a Hydrogen atom with an ionization energy about 1.7 eV
D. The impurity will look like a Hydrogen atom with an ionization energy about 3.4 eV
E. The impurity will look like a Hydrogen atom with an ionization energy about 0.85 eV
F. The Bohr radius of the impurity is about 2.12 Angstrom
G. The Bohr radius of the impurity is about 0.53 Angstrom
H. The Bohr radius of the impurity is about 0.14 Angstrom

Answer 1

(E) The impurity will look like a hydrogen atom with an ionization energy about 0.85 eV

(F) The bohr radius of impurity is about 2.12 Angstrom.

Explanation :

Ionization energy of hydrogen atom is given by

Thus decreasing charge by 2 and increasing dielectric constant by 2, decreases ionization energy by 16 times. (= 13.6 eV/16 = 0.85 eV)

And Bohr radius is

Thus decreasing charge by twice and increasing dielectric constant by 2 increases bohrs radius 4- times. (= 0.53 Angstrom x 4 = 2.12 Angstrom)