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CBSE Class 12 · 2016 · Delhi · Modern PhysicsShort Answer

Calculate the energy in the fusion reaction

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

, where the binding energy of

_{1}^{2} H = 2.23 MeV

and of

_{2}^{3} He = 7.73 MeV

.

Answer: $Q = 3.27 MeV$

Solution

Energy released

=

(total BE of products)

-

(total BE of reactants). The free neutron has zero binding energy.

Q = BE (^{3} He) - 2 BE (^{2} H) = 7.73 - 2 (2.23) = 7.73 - 4.46 = 3.27 MeV .

Q = 3.27 MeV

CBSE Class 12 Board · 2016 · Delhi · Subjective

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