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Physics - Modern Physics Question with Solution | TestHub

PhysicsModern PhysicsNuclear physicsHard3 min

PhysicsHardnumerical

Passage / Comprehension

When an α-particle with kinetic energy K = 14 MeV collides with a stationary lithium nucleus, the following nuclear reaction takes place: ⁴He + ⁷Li → ¹⁰B + n. The Q-value of this reaction is E = −2.9 MeV and the neutron is outgoing at right angle to the incoming direction of the α-particle. Kinetic energies of the neutron and the boron nucleus are K1 MeV and K2 MeV respectively.

The value of $K_{2}$ is ______

Answer:

6.10

Solution:

Momentum (neutron ⊥ α): $10 K_{2} = 4 K + K_{1}$ , so $10 K_{2} - K_{1} = 56$

Energy: $K_{1} + K_{2} = 14 - 2.9 = 11.1$

Solving: $11 K_{2} = 67.1$ , $K_{2} = 6.1$ MeV

Answer 6.10

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 3m

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