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Question: A donor atom in a semiconductor has a loosely bound electron. The orbit of this electron is consider...

A donor atom in a semiconductor has a loosely bound electron. The orbit of this electron is considerably affected by the semiconductor material but behaves in many ways like an electron orbiting a hydrogen nucleus. Given that the electrons has an effective mass of 0.07me0.07\,{{\text{m}}_{\text{e}}} (where me{{\text{m}}_{\text{e}}} is mass of the free electron) and the space in which it moves has a permittivity 13ε013\,{\varepsilon _0}, then the radius of the electron's lowermost energy orbit will be close to (The Bohr radius of the hydrogen atom is 0.53A00.53\,\mathop {\text{A}}\limits^0 ).
A. 0.53A00.53\,\mathop {\text{A}}\limits^0
B. 243A0243\,\mathop {\text{A}}\limits^0
C. 10A010\,\mathop {\text{A}}\limits^0
D. 100A0100\,\mathop {\text{A}}\limits^0

Explanation

Solution

First of all, we will find the proper equation with the help of Bohr postulated formula. After that we will obtain the expression of velocity followed by radius. We will substitute the required values of a hydrogen atom followed by manipulation to obtain the answer.

Formula used:
According to Bohr postulates:
kze2r2=mv2r\dfrac{{kz{e^2}}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}
Where,
mm indicates mass
vv indicates velocity
rr indicates radius
zz indicates atomic number.

Complete step by step answer:
From the Bohr postulates formula:
mv2r=KQ2r2\dfrac{{m{v^2}}}{r} = \dfrac{{K{Q^2}}}{{{r^2}}} …… (1)
And
mvr=nh2π v=e22ε0h×zn mvr = \dfrac{{nh}}{{2\pi }} \\\ \Rightarrow v = \dfrac{{{e^2}}}{{2{\varepsilon _0}h}} \times \dfrac{z}{n} \\\ …… (2)
So,
r=nh2πmvr = \dfrac{{nh}}{{2\pi mv}}
Put the value of vv in this equation,
r = \dfrac{{nh}}{{2\pi \mu \left( {\dfrac{{{e^2}}}{{2{\varepsilon _0}h}}} \right)\left( {\dfrac{z}{n}} \right)}} \\\ = \left( {\dfrac{{{\varepsilon _0}{h^2}}}{{\pi m{e^2}}}} \right)\left( {\dfrac{{{n^2}}}{z}} \right) \\\ = {r_0}\dfrac{{{n^2}}}{z} \\\
For the medium of permittivity
ε=13ε0\varepsilon = 13{\varepsilon _0}
Effective mass m=0.07mem = 0.07\,{{\text{m}}_{\text{e}}}
r=13r00.07n2zr = \dfrac{{13\,{r_0}}}{{0.07}}\dfrac{{{n^2}}}{z}
At ground state
n=1n = 1, assuming like HH atom, z=1z = 1
\Rightarrow r = \dfrac{{13}}{{0.07}}\left( {0.53} \right)\mathop {\text{A}}\limits^0 \\\ \Rightarrow r \approx 100\,\mathop {\text{A}}\limits^0 \\\
Hence, the required answer 100A0100\,\mathop {\text{A}}\limits^0 .

So, the correct answer is “Option D”.

Additional Information:
Bohr's postulates: The hydrogen atom model of Bohr is based on three postulates: ( 11 ) an electron travels in a circular orbit around the nucleus, ( 22 ) the angular momentum of an electron is quantified in the orbit, and ( 33 ) the change in the energy of an electron when it makes a quantum leap from one orbit to another is often followed by a photon's emission or absorption. The Bohr model is semi-classical since it incorporates the electron orbit classical principle (postulate 11 ) with the new quantization concept (postulates 22 and 33 ).

Note:
It is important to remember that the Bohr model is significant because the quantization of electron orbits in atoms was the first model to postulate it. It therefore represents an early quantum theory that gave the development of modern quantum theory a start. To describe atomic states, it introduced the idea of a quantum number.