Chapter 13, Problem 43P

(a)

To determine

Show that for non-relativistic particles $Q$ and $K_{\alpha }$ are related by the expression $Q=K_{\alpha }\left(1+\frac{M_{\alpha }}{M_d}\right)$.

Answer to Problem 43P

The non-relativistic particles $Q$ and $K_{\alpha }$ are related by the expression.

Explanation of Solution

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

  $M_d{v}_d=M_{\alpha }{v}_{\alpha }$        (I)

Here, $M_d$ is the mass of the daughter nucleus, $v_d$ is the velocity of the daughter nucleus, $M_{\alpha }$ is the alpha particle, and $v_{\alpha }$ is the velocity of the alpha particle.

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

  $M_p{c}^2=M_d{c}^2+M_{\alpha }{c}^2+\frac{1}{2}{M}_d{v}_d^2+\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2$        (II)

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

The disintegration energy is given by,

  $\begin{array}{c}Q=\left(M_p-M_d-M_{\alpha }\right)c^2\\ =\frac{1}{2}{M}_d{v}_d^2+\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2\end{array}$        (III)

Rewrite the equation (I) for $v_d$.

  $v_d=\left(\frac{M_{\alpha }}{M_d}\right)v_{\alpha }$        (IV)

Conclusion:

Substitute $v_d$ from the equation (IV) in equation (III).

  $\begin{array}{c}Q=\frac{1}{2}{M}_d{\left[\left(\frac{M_{\alpha }}{M_d}\right)v_{\alpha }\right]}^2+\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2\\ =\frac{1}{2}\left(\frac{M_{\alpha }^2}{M_d}\right)v_{\alpha }^2+\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2\\ =\frac{1}{2}{M}_{\alpha }{v}_{\alpha }^2\left(1+\frac{M_{\alpha }}{M_d}\right)\\ =K_{\alpha }\left(1+\frac{M_{\alpha }}{M_d}\right)\end{array}$

Hence, proved the non-relativistic particles $Q$ and $K_{\alpha }$ relation.

(b)

To determine

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

Answer to Problem 43P

The energy of the alpha particle emitted in the decay of ²²⁶Ra is $4.79\text{MeV}$.

Explanation of Solution

From part (a),

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

  $K_{\alpha }=\frac{Q}{1+\frac{M_{\alpha }}{M}}$        (V)

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

Conclusion:

Substitute $4.87\text{MeV}$ for $Q$, $4$ for $M_{\alpha }$, and $226$ for $M$ in equation (V) to find $K_{\alpha }$.

  $\begin{array}{c}{K}_{\alpha }=\frac{\left(4.87\text{MeV}\right)}{1+\frac{4}{226}}\\ =4.79\text{MeV}\end{array}$

Therefore, The energy of the alpha particle emitted in the decay of ²²⁶Ra is $4.79\text{MeV}$.

(c)

To determine

The kinetic energy of the daughter nucleus.

Answer to Problem 43P

The kinetic energy of the daughter nucleus is $0.08\text{MeV}$.

Explanation of Solution

From part (a),

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

  $K_d=Q-K_{\alpha }$        (VI)

Here, $K_d$ is the energy of the daughter nucleus.

Conclusion:

Substitute $4.87\text{MeV}$ for $Q$ and $4.79\text{MeV}$ for $K_{\alpha }$ in equation (VI) to find $K_d$.

  $\begin{array}{c}{K}_d=4.87\text{MeV}-4.79\text{MeV}\\ =0.08\text{MeV}\end{array}$

Therefore, the kinetic energy of the daughter nucleus is $0.08\text{MeV}$.

(d)

To determine

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

Answer to Problem 43P

The daughter nucleus carries off negligible amount of kinetic energy in beta decay is $2.61\times {10}^{-6}$ and actual beta decay involves another particle and relativistic effects.

Explanation of Solution

For the beta decay of ${}^{210}\text{Bi}$ we have the disintegration energy is,

  $Q=K_{e^{-}}\left(1+\frac{M_{e^{-}}}{M_Y}\right)$

Here, $K_{e^{-}}$ is the kinetic energy of an electron, $M_{e^{-}}$ is the mass of an electron, and

$M_Y$ is the mass of the polonium.

Rewrite the above relation for $K_{e^{-}}$.

  $K_{e^{-}}=\frac{Q}{\left(1+\frac{M_{e^{-}}}{M_Y}\right)}$        (VII)

Conclusion:

Substitute $5.486\times {10}^{-4}\text{u}$ for $M_{e^{-}}$ and $209.982\text{u}$ for $M_Y$ in equation (VII) to find $K_{e^{-}}$.

  $\begin{array}{c}{K}_{e^{-}}=\frac{Q}{\left(1+\frac{5.486\times {10}^{-4}\text{u}}{209.982\text{u}}\right)}\\ =\frac{Q}{1+2.61\times {10}^{-6}}\end{array}$

Consider $2.61\times {10}^{-6}$ for $\epsilon$, so that the above equation becomes,

  $\begin{array}{c}{K}_{e^{-}}=\frac{Q}{1+\epsilon }\\ =Q{\left(1+\epsilon \right)}^{-1}\\ \simeq Q\left(1-\epsilon \right)\\ =Q\left(1-2.61\times {10}^{-6}\right)\end{array}$

The above condition is for $\epsilon \ll 1$. This means the daughter Po carries off only three millionths of the kinetic energy available in the decay.