wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A copper wire of mass 20g carries a current of 6.023A. What fraction of the total number of free electrons present in the copper wire flows through the cross-section of the copper wire in 2sec? (gram atomic weight of copper is 63.5g).


Open in App
Solution

Step 1: Given data and formula:

Given,

M=63.5gm=20gi=6.023At=2sec

We know,

q=it

Where, q is the charge, i is current and t is the time.

Step 2: Finding the number of electrons in copper wire:

By substituting the values of current and time, we get,

q=6.023A×2s⇒q=12.046C

Now, the total number of free electrons will be,

q=ne

Where n is the number of electrons and e is the charge on a single electron, we get

n=12.046C1.6×10-19C=7.53×1019

Step 3: Calculating the fraction of free electrons flowing in the copper wire:

We know that according to the mole concept,

mole=mM=NCuNA

Where m is the given mass of copper and M is the molar mass of Copper.

NCu is the number of atoms of Cu.

NA is Avogadro number.

By substituting all the values we get,

NCu=2063.5×6.023×1023

NCu=1.9×1023

The no. of electrons present in one Cu atom is 29.

The total no. of electrons would be,

N=29×1.9×1023N=5.5×1024

Now, the fraction of free electrons would be

nN=7.53×10195.5×1024nN=1.36×10-5


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Drift Velocity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon