Home
Class 12
PHYSICS
Find free electrons per unit volume in a...

Find free electrons per unit volume in a wire of density `10^(4)kg//m^(3)`, atomic mass number 100 and number of free electron per atom is one.

Text Solution

Verified by Experts

Number of free charge particle per unit volume
`(n)=("total free charge particle")/("total volume")`
`:.` Number of free electron per atom means total free electrons = total number of atoms.
`=N_A/M_WxxM`
So `n=(N_A/M_WxxM)/V=N_A/M_Wxxd`
`=(6.053xx10^23xx10^4)/(100xx10^-3)`
`n=6.023xx10^28 m^-3`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    MOTION|Exercise EXERCISE -1 (SECTION -A Definition of Current , Current Density , Drift Velocity )|8 Videos

  • CURRENT ELECTRICITY

    MOTION|Exercise EXERCISE -1(SECTION B- Resistance & resistivity , Ohms law +VI (Volt - curr.) Characteristics + Colour Code + Temp . Dependence|8 Videos

  • CURRENT ELECTRICITY

    MOTION|Exercise EXERCISE -3 SECTION -B (PREVIOUS YEAR PROBLEMS JEE MAIN)|42 Videos

  • CONSTRAINED MOTION

    MOTION|Exercise EXAMPLES|12 Videos

  • ELASTICITY

    MOTION|Exercise EXERCISE -3|60 Videos

Similar Questions

Explore conceptually related problems

We know that copper is a conductor of electricity due to presence of free electrons in it. Assume that on an average each copper atom provides one free electron. Find the number of free electrons per unit valume of copper. The density of copper is approximately 9,000 kg / m^3 and its atomic mass is 63.5 u.

An electric current of 16 A exists in a metal wire of cross section 10^(-6)"M"^(2) and length 1m. Assuming one free electron per atom. The drift speed of the free electron in the wire will be (0.00x)m//s. Find value of x. ("Density of metal"= 5xx10^(3)"kg"//"m"^(3), "atomic weight"=60)

Knowledge Check

  • In diamond, the number of free electrons are:

    A
    1
    B
    2
    C
    0
    D
    4

  • There is a current of 40 ampere in a wire of 10^(-6)m^(2) are of cross-section. If the number of free electron per m^(3) is 10^(29) then the drift velocity will be

    A
    `1.25 xx 10^(3) m//s`
    B
    `2.50 xx 10^(-3)m//s`
    C
    `25.0 xx 10^(-3)m//s`
    D
    `250 xx 10^(-3)m//s`

  • How many number of free electrons present on each carbon atoms in graphite

    A
    zero
    B
    3
    C
    2
    D
    1

    • Similar Questions

    Explore conceptually related problems

    Calculate the relaxation time and mean free path at room temperature (i.e. 27^(@)C ). If the number of free electrons per unit volume is 8.5 xx 10^(28)//m^(3) and resistivity rho = 1.7 xx 10^(8) Omega-m . Given that mass of electron = 9.1 xx 10^(-31) kg e = 1.6 xx 10^(-19)C and k = 1 .36 xx 10^(-23) JK^(-1)

    An electric current of 16 A exists in a metal wire of cross section 10^(-6) m^(2) and length 1 m. Assuming one free electron per atom. The drift speed of the free electrons in the wire will be (Density of metal = 5 xx 10^(3) kg//m^(3) atomic weight = 60)

    Quantum number 'm' of a free gaseous atom is associated with

    The number density of free electrons in the semiconductor is 10^(18)m^(-3) . It is doped with a pentavalent impurity atoms of number density 10^(24)m^(-3) , the number density of free electrons m^(-3) increases by a factor of

    The number of free electrons per unit volume in a material is n and when electric field E is applied, these electrons are found to aquire a drift velocity of v_(d) . What will be the resistivity of the material?

    Home
    Profile