Answer:
(a) [tex]2.3125\times 10^{10}\ electrons[/tex]
(b) [tex]9.577\times 10^{-13}\ electrons/atom[/tex]
Explanation:
(a) Let the number of the excess electrons on sphere is x.
The charge of the electron (e) = [tex]-1.60\times 10^{-19}\ C[/tex].
Given, the net charge (q) = [tex]-3.70\times 10^{-9}\ C[/tex]
Number of excess electrons is:
[tex]n=\frac {q}{e}=\frac {-3.70\times 10^{-9}\ C}{-1.60\times 10^{-19}\ C}=2.3125\times 10^{10}\ electrons[/tex]
(b) Given that the mass of the sphere = 8.30 g
Molar mass = 207 g/mol
Since, 1 mole of Pb contains 6.022×10²³ atoms of Pb
Also, 1 mole mass = 207 g
207 g of Pb contains 6.022×10²³ atoms of Pb
1 g of Pb contains 6.022×10²³ / 207 atoms of Pb
So,
8.30 g of Pb contains (6.022×10²³ / 207)*8.30 atoms of Pb
No of atoms of Pb in 8.30g = 2.4146×10²² atoms
[tex]Excess\ electrons\ per\ lead\ atom=\frac {2.3125\times 10^{10}\ electrons}{2.4146\times 10^{22} atoms}=9.577\times 10^{-13}\ electrons/atom[/tex]
