Chapter 29, Problem 38P

(a)

To determine

The mass of the particles on the left side.

Answer to Problem 38P

The mass of the particles on the left side is 8.023829 u.

Explanation of Solution

The mass of the particles on the left side is,

$m_i=m\left(\text{H}\right)+m\left(\text{L}\text{i}\right)$

• $m\left(\text{H}\right)$ is the mass of $\text{H}$.
• $m\left(\text{L}\text{i}\right)$ is the mass of $\text{L}\text{i}$.

Substitute 1.007825 u for $m\left(\text{H}\right)$ and 7.016004 u for $m\left(\text{L}\text{i}\right)$.

$\begin{array}{c}{m}_i=\left(\text{1}\text{.007825 u}\right)+\left(\text{7}\text{.016004 u}\right)\\ =8.023829\text{ u}\end{array}$

Conclusion:

The mass of the particles on the left side is 8.023829 u.

(b)

To determine

The mass of the particles on the right side.

Answer to Problem 38P

The mass of the particles on the right side is 8.025594 u.

Explanation of Solution

The mass of the particles on the left side is,

$m_f=m\left(\text{B}\text{e}\right)+m\left(\text{n}\right)$

• $m\left(\text{B}\text{e}\right)$ is the mass of $\text{B}\text{e}$.
• $m\left(\text{n}\right)$ is the mass of $\text{n}$.

Substitute 7.016929 u for $m\left(\text{B}\text{e}\right)$ and 1.008665 u for $m\left(\text{n}\right)$.

$\begin{array}{c}{m}_f=\left(\text{7}\text{.016929 u}\right)+\left(\text{1}\text{.008665}\text{u}\right)\\ =8.025594\text{ u}\end{array}$

Conclusion:

The mass of the particles on the right side is 8.025594 u.

(c)

To determine

The Q-value.

Answer to Problem 38P

The Q-value is $-1.64\text{MeV}$.

Explanation of Solution

Formula to calculate the Q-value is,

$Q=\left(m_i-m_f\right)c^2$

• c is the speed of light.

Substitute 8.023829 u for $m_i$, 8.025594 u for $m_f$ and 931.5 MeV/u for $c^2$.

$\begin{array}{c}Q=\left[\left(\text{8}\text{.023829 u}\right)-\left(\text{8}\text{.025594 u}\right)\right]\left(\text{931}\text{.5}\text{MeV/u}\right)\\ =-1.64\text{MeV}\end{array}$

Conclusion:

The Q-value is $-1.64\text{MeV}$.

(d)

To determine

The expression describing the law of conservation of momentum.

Answer to Problem 38P

The expression describing the law of conservation of momentum is $m_{p}v=\left(m_{Be}+m_n\right)V$

Explanation of Solution

Formula to calculate the initial momentum is,

$p_i=m_{p}v$      (I)

Formula to calculate the final momentum is,

$\begin{array}{c}{p}_f=m_{Be}V+m_{n}V\\ =\left(m_{Be}+m_n\right)V\end{array}$      (II)

From Law of conservation of momentum,

$p_i=p_f$      (III)

Substitute Equations (I) and (II) in (III).

$m_{p}v=\left(m_{Be}+m_n\right)V$

Conclusion:

The expression describing the law of conservation of momentum is $m_{p}v=\left(m_{Be}+m_n\right)V$

(e)

To determine

The expression relating the kinetic energies of the particles.

Answer to Problem 38P

The expression relating the kinetic energies of the particles is $\frac{1}{2}{m}_p{v}^2+Q=\frac{1}{2}\left(m_{Be}+m_n\right)V^2$

Explanation of Solution

Formula to calculate the initial kinetic energy is,

$K{E}_i=\frac{1}{2}{m}_p{v}^2+Q$      (IV)

Formula to calculate the final kinetic energy is,

$K{E}_f=\frac{1}{2}\left(m_{Be}+m_n\right)V^2$      (V)

From Law of conservation of energy,

$K{E}_i=K{E}_f$      (VI)

Substitute Equations (IV) and (V) in (VI).

$\frac{1}{2}{m}_p{v}^2+Q=\frac{1}{2}\left(m_{Be}+m_n\right)V^2$

Conclusion:

The expression relating the kinetic energies of the particles is $\frac{1}{2}{m}_p{v}^2+Q=\frac{1}{2}\left(m_{Be}+m_n\right)V^2$

(f)

To determine

The minimum kinetic energy of the proton.

Answer to Problem 38P

The minimum kinetic energy of the proton is 1.88 MeV.

Explanation of Solution

Formula to calculate the minimum kinetic energy of the proton is,

$K{E}_{\min }=\left(1+\frac{m\left(\text{H}\right)}{m\left(\text{L}\text{i}\right)}\right)\left|Q\right|$

Substitute 1.007825 u for $m\left(\text{H}\right)$, 7.016004 u for $m\left(\text{L}\text{i}\right)$ and $-1.64\text{MeV}$ for $Q$.

$\begin{array}{c}K{E}_{\min }=\left(\text{1+}\frac{\text{1}\text{.007825 u}}{\text{7}\text{.016004 u}}\right)\left|-1.64\text{MeV}\right|\\ =\left(\text{1+}\frac{\text{1}\text{.007825 u}}{\text{7}\text{.016004 u}}\right)\left(1.64\text{MeV}\right)\\ =1.88\text{MeV}\end{array}$

Conclusion:

The minimum kinetic energy of the proton is 1.88 MeV