Question

Excess electrons are placed on a small lead sphere with a mass of 8.40 g so that its net charge is −3.70×10−9 C. Find the number of excess electrons on the sphere and the number of excess electrons per lead atom.

Answer 1

As you know

Q=ne

here Q=-3.7×10^(-9) C , e=-1.6×10^(-19)

so number of excess electron on the sphere

. n=Q/e =-3.7×10^(-9)/(-1.6×10^(-19))

n=2.3125×10^10 Answer

Since 207.2g lead has number of atom =6.023×10^23

So 8.4g has .=6.023×10^23 ×8.4/207.2 = 0.2442×10^23

= 2.442×10^22 atoms

So number of excess electron per lead atom

=2.3125×10^10/2.442×10^22

=0.947×10^(-12)

=9.47×10^(-13). Answer.