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CBSE Class 12 · 2024 · Modern PhysicsShort Answer

Calculate the energy released / absorbed (in MeV) in the nuclear reaction

_{1}^{1} H +_{1}^{3} H \to_{1}^{2} H +_{1}^{2} H

. Given

m (_{1}^{1} H) = 1.007825 u

,

m (_{1}^{2} H) = 2.014102 u

,

m (_{1}^{3} H) = 3.016049 u

.

Answer: $Q \approx - 4.03 MeV$ (energy absorbed).

Solution

Mass change (products

-

reactants):

Δ m = 2 (2.014102) - (1.007825 + 3.016049) = 4.028204 - 4.023874 = + 0.004330 u .

Since the products are heavier than the reactants, energy is absorbed:

Q = - Δ m \times 931.5 = - 0.004330 \times 931.5 \approx - 4.03 MeV .

Q \approx - 4.03 MeV (energy absorbed)

CBSE Class 12 Board · 2024 · Subjective

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