Question
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
a) Find the number of excess electrons on the lead sphere
b) How many electrons are there per lead atom
Answers
(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
(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
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