Answer
See the detailed answer below.
Work Step by Step
We have the decay of:
$$
n \rightarrow p^+ + e^- + \bar{\nu}_e
$$
$$\color{blue}{\bf [a]}$$
We need to find the Total Kinetic Energy of the proton and electron.
$$K=\Delta m c^2$$
The neutron has a mass slightly greater than the combined mass of the proton and electron.
Where:
- $ m_n = 1.00866 \, \text{u} $
- $ m_p = 1.00728 \, \text{u} $
- $ m_e = 0.00055 \, \text{u} $
$$K=( m_n - m_p - m_e)c^2$$
where $ c^2 = 931.49 \, \text{MeV}/\text{u} $
$$
K = (1.00866 - 1.00728 - 0.00055)(931.49 )
$$
$$K=\color{red}{\bf 0.773}\;\rm MeV$$
$$\color{blue}{\bf [b]}$$
The energy conservation equation can be written as:
$$
m_n c^2 = \gamma_p m_p c^2 + \gamma_e m_e c^2
$$
where $ \gamma_p $ and $ \gamma_e $ are the relativistic factors for the proton and electron, respectively whereas $\gamma_n=1$ since the neutron was initially at rest.
Rearrange to express the energy condition:
$$
\boxed{m_n = \gamma_p m_p + \gamma_e m_e}
$$
$$
\boxed{\Delta m= m_n - \gamma_p m_p - \gamma_e m_e}
$$
$$\color{blue}{\bf [c]}$$
Since the neutron is initially at rest, the total momentum before decay is zero, so:
$$
\boxed{\gamma_p m_p v_p = \gamma_e m_e v_e}
$$
$$\color{blue}{\bf [d]}$$
Recalling that
$$
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
$$
Solving for $ v $ gives:
$$
\gamma^2 = \frac{1}{ 1 - \frac{v^2}{c^2}}
$$
$$
1 - \frac{v^2}{c^2}= \frac{1}{ \gamma^2 }
$$
$$
\frac{v^2}{c^2}= 1 - \frac{1}{ \gamma^2 }
$$
$$
v^2 = c^2\left[1 - \frac{1}{ \gamma^2 }\right]
$$
Square root:
$$
v = c\left[ \frac{\gamma^2-1}{ \gamma^2 }\right]^{1/2}
$$
$$
v = \frac{c}{\gamma}\left[ \gamma^2-1\right]^{1/2}
$$
$$
\boxed{v \gamma= c\left[ \gamma^2-1\right]^{1/2}}
$$
This previous boxed formula can help determine $ v_p $ and $ v_e $ once $ \gamma_p $ and $ \gamma_e $ are calculated.
$$\color{blue}{\bf [e]}$$
From the boxed formula in part (c) and the boxed formula in part (d), we get
$$(\gamma_p v_p) m_p= (\gamma_e v_e)m_e $$
$$\color{red}{\bf\not} c\left[ \gamma_p^2-1\right]^{1/2} m_p= \color{red}{\bf\not} c \left[ \gamma_e^2-1\right]^{1/2} m_e $$
Square both sides;
$$ \left[ \gamma_p^2-1\right] m^2_p= \left[ \gamma_e^2-1\right] m^2_e $$
$$ \gamma_p^2m^2_p-m^2_p = \gamma_e^2m^2_e -m^2_e \tag 1$$
From the first boxed formula in part (b) above, $\gamma_e m_e=m_n - \gamma_p m_p $
$$ \gamma_p^2m^2_p-m^2_p =(m_n - \gamma_p m_p )^2 -m^2_e $$
$$ \gamma_p^2m^2_p-m^2_p = m_n^2 -2\gamma_p m_n m_p +\gamma_p^2m^2_p -m^2_e $$
$\gamma_p^2m^2_p$ is a common factor,
$$ -m^2_p = m_n^2 -2\gamma_p m_n m_p -m^2_e $$
Solve for $\gamma_p$
$$ m_n^2 - m^2_e +m^2_p = 2\gamma_p m_n m_p $$
$$\gamma_p =\dfrac{ m_n^2 - m^2_e +m^2_p}{ 2m_n m_p} $$
Where:
- $ m_n = 1.00866 \, \text{u} $
- $ m_p = 1.00728 \, \text{u} $
- $ m_e = 0.00055 \, \text{u} $
$$\gamma_p =\dfrac{ 1.00866^2 - 0.00055 ^2 +1.00728 ^2}{ 2(1.00866 )(1.00728 )} $$
$$\gamma_p=\color{red}{\bf 1.000000788334251} $$
Solving (1) for $\gamma_e$;
$$ \gamma_e^2 =\dfrac{ \gamma_p^2m^2_p-m^2_p +m^2_e }{m^2_e } $$
$$ \gamma_e =\sqrt{\dfrac{ (1.00000078833)^2(1.00728 )^2-(1.00728 )^2 +(0.00055 )^2 }{(0.00055 )^2 } }$$
$$\gamma_e=\color{red}{\bf 2.50764} $$
Solving the boxed formula in part (d) for $v$;
$$ v =\dfrac{ c}{ \gamma}\left[ \gamma^2-1\right]^{1/2} $$
For $v_e$'
$$ v_e =\dfrac{ c}{ \gamma_e}\left[ \gamma_e^2-1\right]^{1/2}=\dfrac{ c}{ 2.50764}\left[ 2.50764^2-1\right]^{1/2} $$
$$v_e=\color{red}{\bf 0.917 }c$$
For $v_p$'
$$ v_p =\dfrac{ c}{ 1.000000788334251}\left[ 1.000000788334251^2-1\right]^{1/2} $$
$$v_p=\color{red}{\bf 0.0012557}c$$
$$\color{blue}{\bf [f]}$$
Now we need to Calculate the Kinetic Energy of the Proton and Electron
For the proton:
$$
K_p = (\gamma_p - 1)m_p c^2 $$
$$K_p= (1.00000078833 - 1)(1.00728) (931.49) \\
= \bf 0.000739667\, \rm {MeV}
$$
For the electron:
$$
K_e = (\gamma_e - 1)m_e c^2 $$
$$K_e= (2.50764- 1)(0.00055) (931.49)\\
=\bf 0.772393\, \rm {MeV}
$$
The total kinetic energy is then
$$K_{\rm tot}= K_p + K_e =0.000739667+ 0.772393
$$
$$K_{\rm tot}=\color{red}{\bf 0.773} \, \rm {MeV} $$
which matches the calculated value from part (a).
$$\color{blue}{\bf [g]}$$
The proton's recoil energy (0.0007 MeV) is very small, indicating it remains bound in the nucleus.
The electron, moving at nearly the speed of light with significant kinetic energy, has enough energy to escape.
The electron’s de Broglie wavelength is on the order of nanometers, much larger than nuclear dimensions, preventing it from being confined within the nucleus.
