Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Recalling that
$$E=K+U\quad\Rightarrow\quad K=E-U$$
We know that
- The total energy of the alpha particle, $ E = 5 \, \text{MeV} $.
- Inside the nucleus, the potential energy $ U_{in} = -60 \, \text{MeV} $.
- Outside the nucleus, the potential energy $ U_{out} = 0 \, \text{MeV} $.
Thus,
$$
K_\text{in} = E - U_{in} = 5.0 -(- 60.0)= \color{red}{\bf 65} \, \text{MeV}
$$
And
$$
K_\text{out} = E -U_{out}= 5.0 -0= \color{red}{\bf 5} \, \text{MeV}
$$
$$\color{blue}{\bf [b]}$$
First, we need to calculate the Speed of the Alpha Particle Inside,
where $ K_\text{in} = \frac{1}{2}mv^2 $, we can solve for $ v $:
$$
v = \sqrt{\frac{2K_\text{in}}{m}} = \sqrt{\frac{2 \times 65 \times 10^6 \times 1.6 \times 10^{-19}}{4 \times 1.661 \times 10^{-27}}}$$
$$ v=\bf 5.60 \times 10^7 \, \rm {m/s}
$$
Now we need to calculate the time between collisions.
The particle oscillates back and forth within the nucleus, covering a distance of $ 15 \, \text{fm} = 15 \times 10^{-15} \, \text{m} $.
- Time for a one-way trip:
$$
\Delta t = \frac{15 \times 10^{-15} \, \text{m}}{5.60 \times 10^7 \, \text{m/s}} \approx 2.7 \times 10^{-22} \, \text{s}
$$
So, the rate at which the particle collides with one side of the potential barrier is:
$$
R = \frac{1}{\Delta t} = \frac{1}{2.7 \times 10^{-22}} $$
$$R =\color{red}{\bf 3.7 \times 10^{21}} \, \text{collisions/s}
$$
$$\color{blue}{\bf [c]}$$
The probability $ P_\text{tunnel} $ for the particle to tunnel through a 20-fm-wide barrier is given by:
$$
P_\text{tunnel} = e^{-2 w \eta}
$$
where $ w = 20 \, \text{fm} $ is the barrier width, and $ \eta $ is the penetration depth into the classically forbidden region, which is given by
$$ \eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}$$
$$\eta = \frac{1.05 \times 10^{-34} \, \text{J s}}{\sqrt{2 \times (4 \times 1.661 \times 10^{-27}) \times (25 \times 10^6 \times 1.6 \times 10^{-19})}} $$
$$\eta = 4.55 \times 10^{-16} \, \text{m} = 0.455 \, \text{fm}
$$
Substitute $ \eta $ and $ w $ into the tunneling probability formula.
$$
P_\text{tunnel} = e^{-2 ( 20) / 0.455 } $$
$$ P_\text{tunnel} =\color{red}{\bf 6.61 \times 10^{-39}}
$$
$$\color{blue}{\bf [d]}$$
The number of collisions $ N $ required for the probability of the particle remaining inside the nucleus to drop to 0.5 is given by:
$$
N = \frac{1 - P_{in}}{P_{tunnel}} = \frac{1-0.50}{6.61 \times 10^{-39}} $$
$$N= 7.56 \times 10^{37} \text{ collisions} $$
Now we can calculate Half-Life $ t_{1/2} $ by using the collision rate $ R $
$$
t_{1/2} = \frac{N}{R} = \frac{7.56 \times 10^{37}}{3.7 \times 10^{21} }
$$
×10^16
$$
t_{1/2} =\bf 2.04324\times 10^{16} \, \rm {s} =\color{red}{\bf 6.5 \times 10^8 } \;\rm yr
$$
