(a) N=q/e=3.2•10⁹/1.6•10⁻¹⁹= 2•10²⁸
(b)
The mass of one lead atom is
m₀=207.2 amu =207.2•1.67•10⁻²⁷=3.46•10⁻²⁵kg
The number of atoms in 8 g of lead
N₀=m/m₀=0.008/3.46•10⁻²⁵ = 2.31•10²³
N/N₀=2•10²⁸/2.31•10²³=8.65•10⁵
Excess electrons are placed on a small lead sphere with mass 8grams so that its net charge is -3.20*10^9C.
a) Find the number of excess electrons on the lead sphere
b) How many electrons are there per lead atom
1 answer
1 answer