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Question: Pure Si at \[500K\]has equal number of electron \[({n_e})\]and hole \[({n_h})\] concentration\[1.5 \...

Pure Si at 500K500Khas equal number of electron (ne)({n_e})and hole (nh)({n_h}) concentration1.5×1016m−31.5 \times {10^{16}}{m^{ - 3}}. Doping by indium increases nh{n_h} to4.5×1022m−34.5 \times {10^{22}}{m^{ - 3}}. The doped semiconductor is of:
(A) n-type with electron concentration ne=5×1022m−3{n_e} = 5 \times {10^{22}}{m^{ - 3}}
(B) p-type with electron concentration ne=2.5×1010m−3{n_e} = 2.5 \times {10^{10}}{m^{ - 3}}
(C) n-type with electron concentration ne=2.5×1023m−3{n_e} = 2.5 \times {10^{23}}{m^{ - 3}}
(D) p-type with electron concentration ne=5×109m−3{n_e} = 5 \times {10^9}{m^{ - 3}}

Explanation

Solution

The concentration of silicon is given in the question, we know that the square of this concentration must be equal to the concentration of holes and electrons.

We are given the increased concentration of holes on doping, and using the formula we can obtain the concentration of doping.

Complete Step-By-Step Solution:
In the question, it is given that pure silicon has an equal number of holes and electrons, and has a concentration of1.5×1016m−31.5 \times {10^{16}}{m^{ - 3}}.
This is a type of intrinsic semiconductor, or un-doped semiconductor.
We know, that in case of pure semiconductor, (in this case Si):

ni2=ne×nh{n_i}^2 = {n_e} \times {n_h}

Where,
ni={n_i}^{} = Concentration of un-doped semiconductor
ne={n_e} = Number of electrons (concentration of electrons)
nh={n_h} = Number of holes (concentration of holes)
From the question, it is stated that on doping with indium, the concentration of holes becomes4.5×1022m−34.5 \times {10^{22}}{m^{ - 3}}.
The concentration of pure semiconductor1.5×1016m−31.5 \times {10^{16}}{m^{ - 3}}.
Now, putting the values in the equation we obtain:

(1.5×1016m−3)2=4.5×1022m−3×ne{(1.5 \times {10^{16}}{m^{ - 3}})^2} = 4.5 \times {10^{22}}{m^{ - 3}} \times {n_e}

On solving the equation, we obtain:

ne=5×109m−3{n_e} = 5 \times {10^9}{m^{ - 3}}

This is the concentration of electrons in the doped semiconductor.
Now, it is evident from the values that:

nh>>ne{n_h} > > {n_e}

This means that the concentration of holes is greater than that of electrons.
Thus it is a type of P-type semiconductor, having excess positive charge carriers.

Hence, option (D) IS CORRECT.

Note:
When concentration of holes is lesser than that of concentration of electrons, it becomes a N-TYPE semiconductor.
P-Type and N-type semiconductor are classified on the basis of the excess numbers of electrons or holes present in it.
In N-type majority charge carriers are negative whereas in case of P-type the majority charge carriers are positive.