91影视

A simple one-dimensional model of the potential-energy well of an alpha particle in a nucleus is shown below

The kinetic energy is $K=E-U$.

The alpha particle has $E=5MeV$whether it is inside or outside the nucleus.

Inside $U=-60MeV$, the kinetic energy is

$$\begin{gathered} K_{in}=5-\left(-60\right) \\ =65MeV \end{gathered}$$

Outside $U=0MeV$, the kinetic energy is

$$\begin{gathered} K_{out}=5-0 \\ =5MeV \end{gathered}$$

Therefore, the kinetic energy of the alpha particle while inside the nucleus is$65MeV$and kinetic energy after it escapes from the nucleus is$5MeV$.

The kinetic energy is given by $K_{in}=\frac{1}{2}m{v}^2$.

The speed of the particle is

$$\begin{gathered} v=\sqrt{\frac{2K_{in}}{m}} \\ =\sqrt{\frac{2脳65脳{10}^6脳1.6脳{10}^{-19}}{4脳1.66脳{10}^{-27}}} \\ =5.60脳{10}^{7}m/s \end{gathered}$$

The time needed to move from one side of the potential well to the other, a distance of 15 fm, is

$$\begin{gathered} 鈭�t=\frac{15脳{10}^{-15}}{5.60脳{10}^7} \\ =2.7脳{10}^{-22}s \end{gathered}$$

The particle collides with one wall or the other of the potential-energy barrier each time it moves across the potential well, so the rate of collisions is

$$\begin{gathered} R=\frac{1}{鈭�t} \\ =\frac{1}{2.7脳{10}^{-22}} \\ =3.7脳{10}^{21}collisions/s \end{gathered}$$

Therefore, the collision rate of particles is $3.7脳{10}^{21}collisions/s$.

The tunneling probability is given by $P_{tunnel}=e^{-\raisebox{1ex}{$2w$}\!\left/ \!\raisebox{-1ex}{$畏$}\right.}$, where $w$is the width of the barrier and 畏 is the penetration distance into the classically forbidden region.

The penetration distance is

$$\begin{gathered} 畏=\frac{魔}{\sqrt{2m\left(U_0-E\right)}} \\ =\frac{1.05脳{10}^{-34}}{\sqrt{2脳4\left(1.66脳{10}^{-27}\right)脳25脳{10}^6脳1.60脳{10}^{-19}}} \\ =4.55脳{10}^{-16}m \\ =0.455fm \end{gathered}$$

The probability of tunneling through the 20-fm-wide barrier is

$$\begin{gathered} P_{tunnel}=e^{-2\left(20\right)\left(0.455\right)} \\ =6.6脳{10}^{-39} \end{gathered}$$

Therefore, the probability of tunneling through the 20-fm-wide barrier is$6.6脳{10}^{-39}$.

Although the probability of tunneling is extremely small, the alpha particle collides with the barrier a very large number of times per second. The probability that the particle is still inside the nucleus after N collisions is $P_{in}=1-N{P}_{tunnel}$.

The number of collisions required to reduce $P_{in}$to $0.50$

$$\begin{gathered} N=\frac{1-P_{in}}{P_{tunnel}} \\ =\frac{0.50}{6.6脳{10}^{-39}} \\ =7.6脳{10}^{37}collisions \end{gathered}$$

At a collision rate of $3.7脳{10}^{21}collisions/s$, the half-life is

$$\begin{gathered} t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{7.6脳{10}^{37}}{3.7脳{10}^{21}} \\ =2.05脳{10}^{16}s脳\frac{1yr}{3.15脳{10}^{21}} \\ =6.5脳{10}^{8}years \\ =650millionyears \end{gathered}$$

Therefore, the half-life of the nucleus is $650millionyears$.