The mass defect in a particular reaction is 0.4 g . The amount of energy liberated is n × 1 0 7 kWh , where n = . (Speed of light c = 3 × 1 0 8 , m s − 1 )

E = Δ m c 2 = 0.4 × 1 0 − 3 × ( 3 × 1 0 8 ) 2 = 3600 × 1 0 7 kW s = 3600 3600 × 1 0 7 kWh = 1 × 1 0 7 kWh

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The mass defect in a particular reaction is $0.4 g$ . The amount of energy liberated is $n \times 1 0^{7} kWh$ , where $n =$ .

(Speed of light $c = 3 \times 1 0^{8}, m s^{- 1}$ )

Answer:

Solution:

$E = Δ m c^{2}$

$= 0.4 \times 1 0^{- 3} \times (3 \times 1 0^{8})^{2}$

$= 3600 \times 1 0^{7} kW s$

$= \frac{3600 \times 1 0 ^{7}}{3600} kWh$

$= 1 \times 1 0^{7} kWh$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: QB

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