Chapter 9, Problem 34P

(a)

To determine

The initial system of the unstable particle.

Answer to Problem 34P

The initial system of the unstable particle is isolated.

Explanation of Solution

Isolated system is system which does not allow the flow of mass energy in our out.

Here, the unstable particle system does not allow any disturbance form the interference of mass-energy from outside the system or didn’t allowed any mass–energy to flow out of the system initially. The fragments of smaller mass of the unstable nuclei are also a part of initial system which further carries out the mass and energy with themselves after the decay but there is no involvement of the mass and energy from outside the system.

Thus the initial system is isolated.

Conclusion:

Therefore, the initial system before the decay is isolated.

(b)

To determine

The analysis model appropriate for analysis of the given isolated system in part (a).

Answer to Problem 34P

The two appropriate analysis models are the conservation of momentum in an isolated system and conservation of energy is an isolated system.

Explanation of Solution

In an isolated system both the total momentum as well as total energy is conserved.

Determine the mass of the fragmented particles using the concept of conservation of total momentum and conservation of total energy.

The single unstable nuclei decayed into two fragments and there is no interference of any type of energy from outside and neither there is any flow of energy from outside the system. Therefore the momentum and the energy will be conserved during and after the fragmentation.

Thus the analysis which can be used to determine the mass of the two fragmented particles conservation of momentum in an isolated system and conservation of energy in an isolated system.

Conclusion:

Therefore, the two appropriate analysis models are the conservation of momentum in an isolated system and conservation of energy is an isolated system.

(c)

To determine

The relativistic factor for the two particles after decay.

Answer to Problem 34P

The values for $\gamma$ for the two particles are $6.22$ and $2.01$.

Explanation of Solution

Given info: The mass of the unstable particle is $3.34\times {10}^{-27}\text{kg}$. The mass of the fragmented particle one is $m_1$, the mass of fragmented particle $m_2$. the velocity of particle with mass $m_1$ is $0.987c$ and the velocity of the particle with mass $m_2$ is $-0.868c$.

The speed of light is $3\times {10}^8{\text{ms}}^{\text{-1}}$.

The formula to calculate the relativistic value is,

$\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$ (1)

Here,

$\gamma$ is the relativistic factor.

$u$ is the velocity for the fragment.

$c$ is the speed of the light.

For first fragment,

Substitute $0.987c$ for $u$ in equation (1).

$\begin{array}{c}{\gamma }_1=\frac{1}{\sqrt{1-\frac{{\left(0.987c\right)}^2}{c^2}}}\\ =6.22\end{array}$

Thus the value of $\gamma$ for fragmented particle on is $6.22$.

For second fragment

Substitute $0.868c$ for $u$ in equation (1).

$\begin{array}{c}{\gamma }_2=\frac{1}{\sqrt{1-\frac{{\left(0.868c\right)}^2}{c^2}}}\\ =2.01\end{array}$

Thus the value of $\gamma$ for fragmented particle second is $2.01$.

Conclusion:

Therefore, value of relativistic factor for fragment one is $6.22$ and for the fragment second is $2.01$.

(d)

To determine

The relation between the masses of two fragments using the conservation of energy analysis.

Answer to Problem 34P

The relation between the mass of fragment one and fragment second is $m_2+3.09m_1=1.66\times {10}^{-27}\text{kg}$.

Explanation of Solution

From the conservation of energy the sum of the energy of the fragments is equal to the sum of the energy of the unstable nuclei.

$E_1+E_2=E_{\text{total}}$ (2)

Here,

$E_1$ is the energy of the $m_1$ fragment.

$E_2$ is the energy of the $m_2$ fragment.

$E_{\text{total}}$ is the total energy.

The formula to calculate the total relativistic energy is,

$E_{\text{total}}=m_{\text{total}}{c}^2$

Here,

$m_{\text{total}}$ is the total mass of the unstable particle.

The formula to calculate relativistic energy of a particle $m_1$ is,

$E_1={\gamma }_1{m}_1{c}^2$

The formula to calculate the energy for particle $m_2$ is,

$E_2={\gamma }_2{m}_2{c}^2$

Substitute ${\gamma }_1{m}_1{c}^2$ for $E_1$, $\gamma m_{\text{total}}{c}^2$ for $E_{\text{total}}$ and ${\gamma }_2{m}_2{c}^2$ for $E_2$ in equation (2).

${\gamma }_1{m}_1{c}^2+{\gamma }_2{m}_2{c}^2=m_{\text{total}}{c}^2$

Substitute $6.22$ for ${\gamma }_1$ and $2.01$ for ${\gamma }_2$ and $3.34\times {10}^{-27}\text{kg}$ for $m_{\text{total}}$ in the above equation.

$\begin{array}{c}\left(6.22\right)m_1{c}^2+\left(2.01\right)m_2{c}^2=\left(3.34\times {10}^{-27}\text{kg}\right)c^2\\ m_2+3.09m_1=1.66\times {10}^{-27}\text{kg}\end{array}$

Thus the relation between the mass of fragment one and fragment second is,

$m_2+3.09m_1=1.66\times {10}^{-27}\text{kg}$

Conclusion:

Therefore, the relation between the mass of fragment one and fragment second is $m_2+3.09m_1=1.66\times {10}^{-27}\text{kg}$.

(e)

To determine

The relation between the masses of two fragments.

Answer to Problem 34P

The relation between the masses $m_1$ and $m_2$ using the conservation of momentum analysis is $m_2=3.52m_1$.

Explanation of Solution

Given info: The mass of the unstable nuclei is $3.34\times {10}^{-27}\text{kg}$. The mass of the fragmented particles is $m_1$ and $m_2$. the velocity of particle $m_1$ is $0.987c$ and the velocity of the particle $m_2$ is $-0.868c$.

From the conservation of momentum the final momentum is zero,

$p_1=p_2$ (3)

Here,

$p_1$ is the momentum of first fragment.

$p_2$ is the momentum of the second fragment.

The momentum of first fragment is,

$p_1={\gamma }_1{m}_1{u}_1$

The momentum of the second fragment is,

$p_2={\gamma }_2{m}_2{u}_2$

Substitute ${\gamma }_2{m}_2{u}_2$ for $p_2$ and ${\gamma }_1{m}_1{u}_1$ for $p_1$ in equation (1).

${\gamma }_1{m}_1{u}_1={\gamma }_2{m}_2{u}_2$

Substitute $6.22$ for ${\gamma }_1$, $0.987c$ for $u_1$, $2.01$ for ${\gamma }_2$ and $0.868c$ for $u_2$ in the above equation.

$\begin{array}{c}{\gamma }_1{m}_1{u}_1={\gamma }_2{m}_2{u}_2\\ \left(6.22\right)\left(0.987c\right)m_1=\left(2.01\right)\left(0.868c\right)m_2\\ m_2=3.52m_1\end{array}$

The relation between the two fragment is $m_2=3.52m_1$.

Conclusion:

Therefore, from the analysis model of conservation of momentum the relation between the two masses is $m_2=3.52m_1$.