Modern Physics
Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 13, Problem 43P

(a)

To determine

Show that for non-relativistic particles Q and Kα are related by the expression Q=Kα(1+MαMd).

(a)

Expert Solution
Check Mark

Answer to Problem 43P

The non-relativistic particles Q and Kα are related by the expression.

Explanation of Solution

According to the conservation of momentum for the masses of daughter nucleus and alpha particle,

  Mdvd=Mαvα        (I)

Here, Md is the mass of the daughter nucleus, vd is the velocity of the daughter nucleus, Mα is the alpha particle, and vα is the velocity of the alpha particle.

According to the conservation of energy for the alpha decay process,

  Mpc2=Mdc2+Mαc2+12Mdvd2+12Mαvα2        (II)

Here, Mp is the mass of the parent nucleus and c is the velocity of light.

The disintegration energy is given by,

  Q=(MpMdMα)c2=12Mdvd2+12Mαvα2        (III)

Rewrite the equation (I) for vd.

  vd=(MαMd)vα        (IV)

Conclusion:

Substitute vd from the equation (IV) in equation (III).

  Q=12Md[(MαMd)vα]2+12Mαvα2=12(Mα2Md)vα2+12Mαvα2=12Mαvα2(1+MαMd)=Kα(1+MαMd)

Hence, proved the non-relativistic particles Q and Kα relation.

(b)

To determine

The energy of the alpha particle emitted in the decay of 226Ra.

(b)

Expert Solution
Check Mark

Answer to Problem 43P

The energy of the alpha particle emitted in the decay of 226Ra is 4.79MeV.

Explanation of Solution

From part (a),

The expression for energy of the alpha particle emitted in the decay of 226Ra is,

  Kα=Q1+MαM        (V)

Here, Q is the disintegration energy and M is the mass of the alpha particle emitted in the decay.

Conclusion:

Substitute 4.87MeV for Q, 4 for Mα, and 226 for M in equation (V) to find Kα.

  Kα=(4.87MeV)1+4226=4.79MeV

Therefore, The energy of the alpha particle emitted in the decay of 226Ra is 4.79MeV.

(c)

To determine

The kinetic energy of the daughter nucleus.

(c)

Expert Solution
Check Mark

Answer to Problem 43P

The kinetic energy of the daughter nucleus is 0.08MeV.

Explanation of Solution

From part (a),

The expression for energy of the daughter nucleus emitted in the decay process is,

  Kd=QKα        (VI)

Here, Kd is the energy of the daughter nucleus.

Conclusion:

Substitute 4.87MeV for Q and 4.79MeV for Kα in equation (VI) to find Kd.

  Kd=4.87MeV4.79MeV=0.08MeV

Therefore, the kinetic energy of the daughter nucleus is 0.08MeV.

(d)

To determine

The daughter nucleus carries off negligible amount of kinetic energy in beta decay and answer an approximation.

(d)

Expert Solution
Check Mark

Answer to Problem 43P

The daughter nucleus carries off negligible amount of kinetic energy in beta decay is 2.61×106 and actual beta decay involves another particle and relativistic effects.

Explanation of Solution

For the beta decay of 210Bi we have the disintegration energy is,

  Q=Ke(1+MeMY)

Here, Ke is the kinetic energy of an electron, Me is the mass of an electron, and

MY is the mass of the polonium.

Rewrite the above relation for Ke.

  Ke=Q(1+MeMY)        (VII)

Conclusion:

Substitute 5.486×104u for Me and 209.982u for MY in equation (VII) to find Ke.

  Ke=Q(1+5.486×104u209.982u)=Q1+2.61×106

Consider 2.61×106 for ε, so that the above equation becomes,

  Ke=Q1+ε=Q(1+ε)1Q(1ε)=Q(12.61×106)

The above condition is for ε1. This means the daughter Po carries off only three millionths of the kinetic energy available in the decay.

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