Question

A copper sphere of mass 2.0 g contains about 2 x 1022 atoms. The charge on the nucleus of
each atom is 29e. If the charge on sphere is +2jIC then fraction of electrons removed
(A) 5.8x102
(B) 1.25 x 1013
(C) 6.28 x 102
(D) 2.16 x 10-11​

Answer 1

Given :-

Mass of copper sphere :- 2g

No of atoms :- 2 × 10^22 atoms

Charge on the nucleus of each atom :- 29e

To Find :-

Fraction of electrons removed if the charge on sphere is 2μC.

Solution :-

We know :-

Q = ne

• Q = charge
• n = number of electrons
• e = electronic charge

$: \sf \implies n = \dfrac{Q}{e}$

$: \sf \implies n = \dfrac{2 \times {10}^{ - 6} }{1.6 \times {10}^{ - 19} }$

$: \sf \implies n = 1.25 \times {10}^{13}$

Fraction of electrons removed :-

$\bf \implies \dfrac{No \: of \: electrons }{Total \: no\: of \: electrons}$

Here :-

Total number of electrons :- 29 × 2 × 10^22

Substitute the values :-

$: \sf \implies \dfrac{1.25 \times {10}^{13} }{29 \times 2 \times {10}^{22} }$

$: \sf \implies 0.0216 \times {10}^{ - 9}$

$: \bf \implies 2.16 \times {10}^{ - 11}$

Fraction of electrons removed = 2.16 × 10^-11

Correct answer :- Option D

Answer 2

Given :-

A copper sphere of mass 2.0 g contains about 2 x 1022 atoms. The charge on the nucleus of  each atom is 29e. If the charge on sphere is +2jIC

To Find :-

Fraction of electron removed

Solution :-

We know that

Charge = number of electrons/electric charge

Charge = 2 × 10⁻⁶/1.6 × 10⁻¹⁹

Charge = 2 × 10⁻⁶ × 10¹⁹/1.6

Charge = 2 × 10⁽¹⁹ ⁻ ⁶⁾/1.6

Charge = 2 × 10¹³/1.6

Charge = 1.25 × 10¹³

Fraction of electron

1.25 × 10¹³/29 × 2 × 10²²

1.25  × 10¹³/58 × 10²²

2.16 × 10⁻¹¹