Chapter 26, Problem 36P

(a)

To determine

Equation for conservation of energy.

Answer to Problem 36P

  The equation for conservation of energy is $K.E_i+m_i{c}^2=K.E_f+m_f{c}^2$.

Explanation of Solution

The formula used to calculate the initial rest energy is,

$E_{R_i}=m_i{c}^2$

• $E_{R_i}$ is the initial rest energy
• $c$ is the speed of light
• $m_i$ is the initial mass

The formula used to calculate the final relativistic energy is,

$\begin{array}{c}{E}_i=K.E_i+E_{R_i}\\ =K.E_i+m_i{c}^2\end{array}$

The formula used to calculate the final rest energy is,

$E_{R_f}=m_f{c}^2$

• $E_{R_f}$ is the final rest energy
• $m_f$ is the final mass

The formula used to calculate the final relativistic energy is,

$\begin{array}{c}{E}_f=K.E_f+E_{R_f}\\ =K.E_f+m_f{c}^2\end{array}$

The formula used to calculate the conservation energy is,

$\begin{array}{c}{E}_i=E_f\\ K.E_i+m_i{c}^2=K.E_f+m_f{c}^2\end{array}$

Thus, the equation for conservation of energy is $K.E_i+m_i{c}^2=K.E_f+m_f{c}^2$.

Conclusion:

The equation for conservation of energy is $K.E_i+m_i{c}^2=K.E_f+m_f{c}^2$.

(b)

To determine

The total mass of initial particle.

Answer to Problem 36P

  The total mass of initial particle is $236.052588\text{u}$.

Explanation of Solution

Given info:

Mass of $U$ is $235.043923\text{ u}$.

Mass of $n$ is $1.008665\text{u}$.

The formula used to calculate the mass of initial particle is,

$m_i=m_U+m_n$

• $m_i$ is the mass of initial particle
• $m_U$ is the mass of $u$
• $m_n$ is the mass of $n$

Substitute $235.043923\text{ u}$ for $m_U$ and $1.008665\text{u}$ for $m_n$ to find $m_i$.

$\begin{array}{c}{m}_i=235.043923\text{ u}+1.008665\text{u}\\ =236.052588\text{u}\end{array}$

Thus, the total mass of initial particle is $236.052588\text{u}$.

Conclusion:

The total mass of initial particle is $236.052588\text{u}$.

 (c)

To determine

The total mass of final particle.

Answer to Problem 36P

  The total mass of final particle is $235.861612\text{u}$.

Explanation of Solution

Given info:

Mass of $La$ is $\text{147}\text{.932236 u}$.

Mass of $Br$ is $\text{86}\text{.9207119 u}$.

The formula used to calculate the mass of initial particle is,

$m_f=m_{La}+m_{Br}+m_n$

• $m_f$ is the mass of initial particle
• $m_{La}$ is the mass of $La$
• $m_{Br}$ is the mass of $Br$

Substitute $\text{147}\text{.932236 u}$ for $m_{La}$ , $\text{86}\text{.9207119 u}$ for $m_{Br}$ and $1.008665\text{u}$ for $m_n$ to find $m_f$.

$\begin{array}{c}{m}_f=\text{147}\text{.932236 u}+\text{86}\text{.9207119 u}+1.008665\text{u}\\ =235.861612\text{u}\end{array}$

Thus, the total mass of final particle is $235.861612\text{u}$.

Conclusion:

The total mass of final particle is $235.861612\text{u}$.

(d)

To determine

The difference between the masses.

Answer to Problem 36P

  The mass difference is $0.190976\text{u}$.

Explanation of Solution

The formula used to calculate the mass difference is,

$\Delta m=m_i-m_f$

• $\Delta m$ is the mass is the mass difference

Substitute $236.052588\text{u}$ for $m_i$ and $235.861612\text{u}$ for $m_f$ to find $\Delta m$.

$\begin{array}{c}\Delta m=236.052588\text{u}-235.861612\text{u}\\ =0.190976\text{u}\end{array}$

Thus, the mass difference is $0.190976\text{u}$.

Conclusion:

The mass difference is $0.190976\text{u}$.

(e)

To determine

The final kinetic energy in MeV.

Answer to Problem 36P

  The final kinetic energy in MeV is $177.893\text{MeV}$.

Explanation of Solution

Consider the initial kinetic energy as zero, the kinetic energy is equal to the mass difference.

$\begin{array}{c}K.E_f=\Delta m\\ =\left(0.190976\text{u}\right)\left(\frac{931.494\text{MeV}}{1\text{u}}\right)\\ =177.893\text{MeV}\end{array}$

Thus, the final kinetic energy in MeV is $177.893\text{MeV}$.

Conclusion:

The final kinetic energy in MeV is $177.893\text{MeV}$