Question

A solid of mass 1 kg has \(6 \times 10^{24}\) atoms. If one electron is removed from every one atom of 0.005% of the atoms, then the charge gained by the solid is

• A. \(+24 \operatorname{C} \)
• B. \(+48 \operatorname{C} \)
• C. \(+96 \operatorname{C} \)
• D. \(+60 \operatorname{C} \)

Hint:

To calculate net charge from electron transfer: 1. Identify the total number of atoms/particles involved. 2. Determine the fraction or percentage of atoms that lose/gain electrons. 3. Calculate the exact number of electrons transferred. 4. Multiply the number of transferred electrons by the elementary charge (\(e = 1.6 \times 10^{-19} \operatorname{C}\)). 5. Determine the sign of the charge: If electrons are removed, the object gains a positive charge. If electrons are added, the object gains a negative charge.

Answer 1

The Correct Option is B

Step 1: Identify the given information.
Total number of atoms in the solid, \( N_{\text{total}} = 6 \times 10^{24} \) atoms.
Percentage of atoms from which an electron is removed = \( 0.005\% \).
One electron is removed from each affected atom.
The charge of a single electron, \( e = 1.6 \times 10^{-19} \, \text{C} \). (This is a fundamental constant.) 
Step 2: Calculate the number of atoms from which electrons are removed.
The number of atoms affected is \( 0.005\% \) of the total number of atoms: \[ N_{\text{affected}} = \frac{0.005}{100} \times N_{\text{total}} \] \[ N_{\text{affected}} = 0.00005 \times (6 \times 10^{24}) \] Convert \( 0.00005 \) to scientific notation: \( 5 \times 10^{-5} \). \[ N_{\text{affected}} = (5 \times 10^{-5}) \times (6 \times 10^{24}) \] \[ N_{\text{affected}} = (5 \times 6) \times (10^{-5} \times 10^{24}) \] \[ N_{\text{affected}} = 30 \times 10^{19} \] \[ N_{\text{affected}} = 3 \times 10^{20} \, \text{atoms} \] Step 3: Calculate the total charge gained by the solid.
When an electron (which carries a negative charge) is removed from an atom, the atom becomes a positive ion. Therefore, the solid gains a net positive charge.
The magnitude of the charge gained by the solid is the number of electrons removed multiplied by the charge of a single electron. \[ Q = N_{\text{affected}} \times e \] \[ Q = (3 \times 10^{20}) \times (1.6 \times 10^{-19} \, \text{C}) \] \[ Q = (3 \times 1.6) \times (10^{20} \times 10^{-19}) \, \text{C} \] \[ Q = 4.8 \times 10^{1} \, \text{C} \] \[ Q = 48 \, \text{C} \] Since electrons are removed, the charge gained by the solid is positive. The final answer is \( \boxed{+48 \, \text{C}} \).