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

Energy liberated is given by, 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 kW h = 1 × 1 0 7 kW h

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PhysicsModern PhysicsNuclear physicsEasy2 minQB

PhysicsEasynumerical

The mass defect in a particular reaction is $0.4 g$ . The amount of energy liberated is $n \times 1 0^{7} kW h$ , where $n =$ _____. (speed of light $= 3 \times 1 0^{8} m s^{- 1}$ )

Answer:

1.00

Solution:

Energy liberated is given by, $E = Δ m c^{2}$

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

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

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

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: QB

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