In Germanium,the concentration of Aluminum is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Aluminum is a trivalent impurity,so it acts as an acceptor,making the semiconductor $p$-type.Given: Intrinsic carrier concentration $n_i = 10^{19}

Vedclass · Accepted Answer

$10^{17} /m^3$

Vedclass · Answer

(A) Aluminum is a trivalent impurity,so it acts as an acceptor,making the semiconductor $p$-type.Given: Intrinsic carrier concentration $n_i = 10^{19} /m^3$ and acceptor concentration $N_A = 10^{21} /m^3$.In a $p$-type semiconductor,the hole concentration $n_h \approx N_A = 10^{21} /m^3$.Using the law of mass action: $n_i^2 = n_h 	imes n_e$.Substituting the values: $(10^{19})^2 = 10^{21} 	imes n_e$.$10^{38} = 10^{21} 	imes n_e$.$n_e = 10^{38} / 10^{21} = 10^{17} /m^3$.Thus,the electron concentration is $10^{17} /m^3$.

In Germanium,the concentration of Aluminum is $\sim 10^{21} \text{ atoms}/m^3$. The intrinsic carrier concentration in the pure semiconductor is $\sim 10^{19} /m^3$. After adding the impurity,what is the electron concentration?

Similar Questions

When phosphorus and antimony are mixed in germanium,then:

When a small amount of impurity atoms are added to a semiconductor,then generally its resistivity:

What will be the conductivity of a pure silicon crystal at $300 \ K$ temperature,if the electron-hole pairs per $cm^3$ is $1.072 \times 10^{10}$ at this temperature,$\mu_e = 1350 \ cm^2/V \cdot s$ and $\mu_h = 480 \ cm^2/V \cdot s$?

$A$ pure semiconductor crystal has $8 \times 10^{28} \text{ atoms/m}^3$. It is doped with a $2 \text{ ppm}$ concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is (Intrinsic carrier concentration,$n_i = 1 \times 10^{16} \text{ m}^{-3}$).

$A$ $P$-type semiconductor has acceptor levels $57 \; meV$ above the valence band. The maximum wavelength of light required to create a hole is (Planck's constant $h = 6.6 \times 10^{-34} \; J \cdot s$, speed of light $c = 3 \times 10^8 \; m/s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module