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CBSE Class 12-science Answered

hello mam When a potential difference of 1.5 V is applied across a wire of length 0.2 m and area of  cross-section 0.30 mm^2 , a current of 2.4 A flows through the wire. If the number density of  free electrons in the wire is 8.4 X 10^28 m^-3 , calculate the average relaxation time.
Asked by seeni2005 | 22 Jun, 2021, 16:15: PM
answered-by-expert Expert Answer
Electric field set up across the conductor,
E = V/t = 1.5/0.2 = 7.5 V/m 
Current density in the wire, 
J = I/ A  = 2.4/(0.3 x 10-6)
= 8 x 106 A/m2
 
Current density, 
J = ne2τE/m 
Thus, 
 
Average relaxation time,
τ = mJ/ ne2E 
= 9.1 x 10-31 x 8 x 106/ (8.4 x 1028) x (1.6 x 10-19)2 x 7.5 
= 4.51 x 10-16 s 
Answered by Shiwani Sawant | 22 Jun, 2021, 17:27: PM

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