Question

A sphere of lead of mass 10 g has net charge -25 xx 10^(-9)C.
(i) Find the number of excess electrons on the sphere.
(ii) How many excess electrons are per lead atom ? Atomic number of lead is 82 and its atomic mass is 207 g/mol.

Answer 1

answer : no of excess electrons is 1.5625 × 10¹¹ and number of excess electrons are per lead atom is 5.37 × 10-¹²

given, mass of sphere of lead , m = 10g

net charge on it, Q = -25 × 10^-9 C

(i) number of electrons on the sphere, n = Q/e

= 25 × 10^-9/1.6 × 10^-19

= (25/1.6) × 10^10

= 15.625 × 10^10

= 1.5625 × 10¹¹

(ii) no of mole = weight in gram/atomic mass

= 10/207 = 0.0483

no of atoms in 10g lead sphere = no of mole × Avogadro's number

= 0.0483 × 6.023 × 10²³

= 0.291 × 10²³

= 2.91 × 10²²

now number of excess electrons per lead = no of electrons/ no of lead atoms

= 1.5625 × 10¹¹/2.91 × 10²²

= 0.537 × 10-¹¹

= 5.37 × 10-¹²