Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

A copper atom has 29 electrons revolving around the nucleus. A copper ball contains 4×1023 atoms. What fraction of the electron be removed to give the ball a charge of +9.6μC?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

~1.8×10-13

b

~1.3×10-12

c

6×10-10

d

~5.2×10-12

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here , No of electrons = 29 

Total no of atoms = 4 x 10 ²³

⇒ Total number of electrons ( N) = 29 x 4 x 10²³ = 116 x 10²³

Charge of the ball (q)= 9.6 μ C the number of electrons = 6x1013

fraction to be removed is 6x 1013116x 10235.2 x10-12

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon