Question

The number of electrons that must be removed from a piece of metal to give it a charge of \( 1\times10^{-7} \) coulomb will be:

• A. \( 10^{7} \)
• B. \( 1.6\times10^{19} \)
• C. \( 6.25\times10^{11} \)
• D. \( 9\times10^{12} \)

Hint:

Charge is quantised: use \( n = Q/e \) with \( e = 1.6\times10^{-19} \) C.

Answer 1

The Correct Option is C

Step 1: Think in terms of charge per electron.
Each electron carries a charge of magnitude \(e = 1.6\times10^{-19}\) C. To build up a total charge of \(Q = 1\times10^{-7}\) C we simply count how many such packets fit into \(Q\).

Step 2: Set up the division.
Number of electrons \(= \dfrac{\text{total charge}}{\text{charge of one electron}} = \dfrac{1\times10^{-7}}{1.6\times10^{-19}}\).

Step 3: Handle the powers of ten first.
\(10^{-7}\div10^{-19} = 10^{12}\). The numerical coefficient is \(1\div1.6 = 0.625\).

Step 4: Combine.
\(0.625\times10^{12} = 6.25\times10^{11}\) electrons.

Step 5: Verify by reverse multiplication.
\(6.25\times10^{11}\times1.6\times10^{-19} = 10\times10^{-8} = 1\times10^{-7}\) C, which is exactly the required charge. The answer is confirmed.

\[\boxed{n = 6.25\times10^{11}}\]