   Chapter 10, Problem 89CP ### University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185

#### Solutions ### University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185
Textbook Problem

# For the reaction, sw n + 3 H e → 4 H e + γ , find the amount of energy transfers to 4 H e and γ (oil the right side of the equation). Assume the reactants are initially at rest. (Hint: Use conservation of momentum principle.)

To determine

The amount of energy transfer to 4He and γ on the right side of equation.

Explanation

Given:

The reaction is, n+4He4He+γ.

Formula used:

The general equation of that nuclide follows is given by,

01n+23He24He+γ

Here, γ is energy release, 01n is neutron.

The mass defect is given by,

Δm=mn+m23He+m24He

Here, Δm is mass defect, mn is mass of neutron, m23He is mass of parent nuclide and m24He is mass of daughter nuclide.

The energy release is given by,

E=(Δm)c2

Here, Δm is change in mass, E is energy and c is speed of light.

The momentum is given by,

p=mv

The equation for energy and momentum is given by,

E=12mv2+|p|c=12mv2+mvc

The energy for γ is given by,

Eγ=|p|c=mvc

Here, Eγ is energy release by γ.

The energy for 4He is given by,

E4He=EEγ

Here, E4He is energy by 4He.

Calculation:

Hence, the atomic mass of parent nuclide helium is, m23He=3.016029u, the atomic mass of daughter nuclide helium is, m24He=4.002603u and the mass of neutron is, mn=1.008645u.

The mass defect is calculated as,

Δm=1.008645u+3.016029u4.002603u=0.022071u

The energy release is calculated as,

E=(0.022071u)c2=(0.022071u)c2( 931.5 MeV/ c 2 1u)=2.065MeV(1.60× 10 13J/MeV)=3.29×1012J

The equation for energy and momentum of 4He is calculated as,

12mv2+mvcE=0

Divide m2 on both side.

v2+2cv+2Em=0

The value of v is calculated as,

v=2c+ ( 2c ) 2 4( 1 )( 2E m )2(1)=2( c+ c 2 +( 2E m ) )2=c+c2+( 2E m )=(3× 108m/s)+ ( 3× 10 8 m/s )2+( 2( 3.29× 10 12 J )( 1 kg m 2 / s 2 1J ) ( 2

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