Question

The mass defect in a particular nuclear reaction is \(0.3 \, \text{g}\). The amount of energy liberated in kilowatt-hours is (velocity of light \( = 3 \times 10^8 \, \text{m/s}\)).

• A. $1.5\times 10^{6}$
• B. $2.5\times 10^{6}$
• C. $3\times 10^{6}$
• D. $7.5\times 10^{6}$

Hint:

$E = \Delta m c^2$, and $1$ kWh = $3.6 \times 10^6$ J.

Answer 1

The Correct Option is D

To find the amount of energy liberated in a nuclear reaction when there is a mass defect, we can use Einstein's mass-energy equivalence principle, expressed by the equation:

\(E = mc^2\)

• E is the energy liberated (in joules).
• m is the mass defect (in kilograms).
• c is the speed of light in vacuum, which is approximately \(3 \times 10^8 \, \text{m/s}\).

Given that the mass defect \(m = 0.3 \, \text{g} = 0.3 \times 10^{-3} \, \text{kg}\), we substitute the values into the formula:

\(E = (0.3 \times 10^{-3} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2\)

Calculating this gives:

\(E = 0.3 \times 10^{-3} \times 9 \times 10^{16} \,\text{J}\)

\(E = 2.7 \times 10^{13} \,\text{J}\)

To convert the energy from joules to kilowatt-hours (kWh), we use the conversion factor:

\(1 \, \text{kWh} = 3.6 \times 10^6 \, \text{J}\)

Thus:

\(\text{kWh} = \frac{2.7 \times 10^{13}}{3.6 \times 10^6}\)

Calculating this gives:

\(\text{kWh} = 7.5 \times 10^{6}\)

Therefore, the amount of energy liberated in the reaction is 7.5 million kilowatt-hours.

Hence, the correct answer is:

$7.5 \times 10^{6}$