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Question: The number of electrons that must be removed from an electrically neutral silver dollar to give it a...

The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of + 2.4C2.4Cis

A

2.5×10192.5 \times 10^{19}

B

1.5×10191.5 \times 10^{19}

C

1.5×10−191.5 \times 10^{- 19}

D

2.5×10−192.5 \times 10^{- 19}

Answer

1.5×10191.5 \times 10^{19}

Explanation

Solution

: Charge on single electron is

e=1.6×10−19Ce = 1.6 \times 10^{- 19}CTotal charge,

q= + 2.4 C Then by quantization of charge q= ne

∴\therefore number of electrons, n=qen = \frac{q}{e}

=2.4C1.6×10−19C=1.5×1019= \frac{2.4C}{1.6 \times 10^{- 19}C} = 1.5 \times 10^{19}