The number of silicon atoms per `m^(3) is 5xx10^(28)`. This is doped simultaneously with `5xx10^(22)` atoms per `m^(3)` of Arsenic and `5xx10^(20) per m^(3)` atoms of indium. Calculate the number of electrons and holes. Given that `n_(i)=1.5xx10^(16)m^(-3)`. Is the material n-type or p-type?
Text Solution
AI Generated Solution
To solve the problem step by step, we will calculate the number of electrons and holes in the doped silicon and determine whether the material is n-type or p-type.
### Step 1: Identify the given values
- Number of silicon atoms, \( N_{Si} = 5 \times 10^{28} \, m^{-3} \)
- Number of arsenic (n-type dopant) atoms, \( N_{As} = 5 \times 10^{22} \, m^{-3} \)
- Number of indium (p-type dopant) atoms, \( N_{In} = 5 \times 10^{20} \, m^{-3} \)
- Intrinsic carrier concentration, \( n_i = 1.5 \times 10^{16} \, m^{-3} \)
...
Topper's Solved these Questions
ELECTRONIC DEVICES
PRADEEP|Exercise SAMPLE PROBLEM|2 Videos
ELECTRONIC DEVICES
PRADEEP|Exercise CONCEPTUAL PROBLEMS|1 Videos
ELECTROMAGNETIC WAVES
PRADEEP|Exercise II Focus multiple choice question|5 Videos
ELECTROSTATICS
PRADEEP|Exercise ASSERTION-REASON TYPE QUESTIONS|2 Videos
Similar Questions
Explore conceptually related problems
A piece of pure semiconductor of silicn of size 1cm xx 1 cm xx 1 mm is having 5 xx 10^(28) number of atoms per cubic metre. It is doped simultaneously with 5 xx 10^(22) atoms per m^(3) of aresenic adn 5 xx 10^(20) per m^(3) atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is 1.5 xx 10^(16) m^(-3) . Mobility of electron is 3800 cm^(2) V^(-)S^(-1) Ratio of conductivity of doped silicon and pure silicon semiconductor is
A piece of pure semiconductor of silicn of size 1cm xx 1 cm xx 1 mm is having 5 xx 10^(28) number of atoms per cubic metre. It is doped simultaneously with 5 xx 10^(22) atoms per m^(3) of aresenic adn 5 xx 10^(20) per m^(3) atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is 1.5 xx 10^(16) m^(-3) . Mobility of electron is 3800 cm^(2) V^(-)S^(-1) The number of holes in this semiconductor are
A piece of pure semiconductor of silicn of size 1cm xx 1 cm xx 1 mm is having 5 xx 10^(28) number of atoms per cubic metre. It is doped simultaneously with 5 xx 10^(22) atoms per m^(3) of aresenic adn 5 xx 10^(20) per m^(3) atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is 1.5 xx 10^(16) m^(-3) . Mobility of electron is 3800 cm^(2) V^(-)S^(-1) The conductivity of doped semiconductor (in Sm^(-1)) is
A piece of pure semiconductor of silicn of size 1cm xx 1 cm xx 1 mm is having 5 xx 10^(28) number of atoms per cubic metre. It is doped simultaneously with 5 xx 10^(22) atoms per m^(3) of aresenic adn 5 xx 10^(20) per m^(3) atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is 1.5 xx 10^(16) m^(-3) . Mobility of electron is 3800 cm^(2) V^(-)S^(-1) The number of electrons in this semiconductor are
Suppose a pure Si-crystal has 5xx10^(28) "atoms" m^(-3) . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Give that n_(i)=1.5xx10^(16)m^(-3) .
Suppose a pure Si crystal has 6xx10^(28) atoms m^(-3) . It is doped by 1ppm concentration of pentavalent As. Calculate the number of electrons and holes, Given that n_(i)=0.5xx10^(16)m^(-3)
Pure Si crystal at 300 k has 2.5 xx10^(28) atoms m^(-3) it is doped by 1 ppm concentration of pentavalent element As calculate the new concentration of electrons and holes take n_(i) =1.5 xx10^(16) m^(-3)
A pure semiconductor of volume 3.5 xx10^(-7) m^(3) contains 5xx10^(20) conduction electrons per m^(3) determine the number of holes present in it
