   Chapter 32, Problem 40PE

Chapter
Section
Textbook Problem

Integrated ConceptsFind the amount of energy given to the 4He nucleus and to the γ ray in the reaction n + 3 He → 4 He + γ , using the conservation of momentum principle and taking the reactants to be initially at rest. This should confirm the contention that must of the energy goes to the γ ray.

To determine

The amount of energy given to the H4e nucleus and to the γ ray in the reaction n+H3eH4e+γ , using the principle of conservation of momentum.

Explanation

Given info:

The equation of the reaction:

n+H3eH4e+γ

From the standard tables:

Mass of neutron

mn=1.008664 u

Mass of H3emH3e=3.016029 u

Mass of H4emH4e=4.002602 u=6.64424×1027kg

Formula used:

Energy released in a nuclear reaction is given by,

E=Δm×931.5 MeV

Here, Δm is the mass defect given by,

Δm=mn+mH3emH4e

Momentum of H4e is given by the expression,

pH4e=2m H 4 eE H 4 e

Here, EH4e is the energy carried away by the H4e.

Momentum of the γ ray is given by

pγ=Eγc

Here, Eγ is the energy carried away by the γ ray.

Calculation:

Calculate the mass defect in the reaction.

Δm=mn+mH3emH4e=1.008664 u+3.016029 u4.002602 u=2.20913×102u

The mass defect is converted into energy, which is carried away by the products of the reaction.

Calculate the energy released in the reaction.

E=Δm×931.5 MeV=2.20913× 10 2u931.5 MeV/u=20.578 MeV

Express the energy in joules.

E=20.578 MeV×1.60× 10 13J1 MeV=3.29248×1012J

This energy is shared between the products H4e nucleus and the γ ray.

Therefore,

E=EH4e+Eγ

..........(1)

The reactants are at rest initially, hence the total initial momentum of the reactants is zero. Since no external force acts on the system, the momentum of the system is conserved. Hence, the final momentum of the system also adds up to zero.

Therefore,

pH4e+pγ=0

And,

pH4e=pγ

The momenta of the H4e nucleus and the γ ray are equal in magnitude but opposite in direction

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