Find the relaxation time for free electrons in copper, if the density of mobile electrons is `8.4 xx10^(28) m^(-3)`. The resistivity of copper at room temperature is `1.7xx10^(-8)Omega m`. Given : mass of electron `=9.11 xx10^(-31) kg ` and charge on electron `=1.6xx10^(-19)C`.
Text Solution
Verified by Experts
Here, `n=8.4xx10^(28) m^(-3)` `rho =1.7 xx 10^(-8) Omega m, m=9.11 xx 10^(-31) kg`, `e=1.6 xx 10^(-19) C, tau=?` We know that, `rho=m/e^(2) n tau` or `tau = m/e^(2) n rho` `(9.11 xx10^(-31))/((1.6 xx 10^(-19))^(2) xx 8.4 xx 10^(28) xx 1.7 xx 10^(-8))` `=2.49 xx10^(-14)s`
Topper's Solved these Questions
CURRENT ELECTRICITY
PRADEEP|Exercise Conceptual Problems|3 Videos
CURRENT ELECTRICITY
PRADEEP|Exercise Very short Q/A|7 Videos
COMMUNICATION SYSTEMS
PRADEEP|Exercise MODEL TEST PAPER-2|9 Videos
DUAL NATURE OF RADIATION AND MATTER
PRADEEP|Exercise Exercise|191 Videos
Similar Questions
Explore conceptually related problems
If the free electron density of copper is 8.6xx10^(28)m^(-3) and resistivity of copper at room temperature is 1.7xx10^(-6)Omega cm , find the relaxation time for the free electrons of copper. Given, mass of electron =9.1xx10^(-31) kg , and charge of electron =1.6xx10^(-19)C .
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 X- ray tube operates at 27 kV . Find the maximum speed of an electron striking the anticathode. Given mass of electron =9xx10^(31)kg and charge on electron =1.6xx10^(-19)C .
An electron is moving with a velocity of 10^(7) m s^(-1) . Find its energy in electron volts. (Mass of electron =9.1xx10^(-31) kg and charge of electron =1.6xx10^(-19) C )
The voltage across the electrodes of a cathode ray tube is 1000 V . Calculate the speed of the electron. Given mass of an electron =.1xx10^(-31)kg and charge on electron =1.6xx10^(16)C .
The mass of half mole of electrons is about (Given: Mass of electron =9.1 xx10^(-28) g)
Find the time of relation between collision and free path of electrons in copper at room temperature .Given resistance of copper = 1.5xx 10^(-8) Omega m number density of electron in copper = 8.5 xx 10^(28)m^(-3) charge on electron = 1.6 xx 10^(19)C , mass of electrons = 9.1 xx 10^(-19) kg Drift velocity of free electron = 1.6 xx 10^(-4)ms^(-1)
Electron density in a copper wire is 8.0 xx10^(22) cm ^(-3) . Calculate the relaxation time for electrons if the resistivity of copper is 1.68xx10^(-8) Omega m at room temperature. (Massof electron is 9.1 xx10^(-31) kg .)
PRADEEP-CURRENT ELECTRICITY-Problems for Practice (B)
