Question:
The amount of energy released when one microgram of matter is annihilated is
The amount of energy released when one microgram of matter is annihilated is
Show Hint
Always convert mass to kg ($1 \text{ \mu g} = 10^{-9} \text{ kg}$) before using $E=mc^2$.
Updated On: Apr 10, 2026
- 25 kWh
- $9 \times 10^{10}$ kWh
- $3 \times 10^{10}$ kWh
- $0.5 \times 10^5$ kWh
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The Correct Option is A
Solution and Explanation
Step 1: Formula
Use Einstein's mass-energy equivalence: $E = mc^2$.
Step 2: Values
$m = 1 \text{ \mu g} = 10^{-9} \text{ kg}$.
$c = 3 \times 10^8 \text{ m/s}$.
$E = 10^{-9} \times (3 \times 10^8)^2 = 9 \times 10^7 \text{ J}$.
Step 3: Conversion
$1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$.
$E = \frac{9 \times 10^7}{3.6 \times 10^6} = 25 \text{ kWh}$.
Final Answer: (a)
Use Einstein's mass-energy equivalence: $E = mc^2$.
Step 2: Values
$m = 1 \text{ \mu g} = 10^{-9} \text{ kg}$.
$c = 3 \times 10^8 \text{ m/s}$.
$E = 10^{-9} \times (3 \times 10^8)^2 = 9 \times 10^7 \text{ J}$.
Step 3: Conversion
$1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$.
$E = \frac{9 \times 10^7}{3.6 \times 10^6} = 25 \text{ kWh}$.
Final Answer: (a)
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