Question

To obtain a rough estimate of the mean time required for uranium-238 to alpha-decay, let us approximate the combined electrostatic and strong nuclear potential energies by rectangular potential barrier half as high as the actual 35 Mev maximum potential energy. Alpha particles (mass 4 u) of 4.3 Mev kinetic energy are incident. Let us also assume that the barrier extends from the radius of nucleus, 7.4 fm to the point where the electrostatic potential drops to 4.3 Mev (i.e., the classically forbidden region). Because $\mathit{U}\mathit{\alpha }\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{r}\mathbf{)}$, this point is 35/4.3 times the radius of the nucleus, the point at which U(r) is 35 Mev. (a) Use these crude approximations, the method suggested in Section 6.3, and the wide-barrier approximation to obtain a value for the time it takes to decay. (b) To gain some appreciation of the difficulties in a theoretical prediction, work the exercise “backward” Rather than assuming a value for U₀, use the known value of the mean time to decay for uranium-238 and infer the corresponding value of U₀, Retain all other assumptions. (c) Comment on the sensitivity of the decay time to the height of the potential barrier.

Short answer

(a)At one decay every10⁵² seconds, the mean lifetime will be 10⁵² seconds or 10⁴⁴ years.

(b)$U_0=1.3\times {10}^{-12}J=7.9MeV$

(c)The decay time is very sensitive with respect to the height of potential barrier. A change of about 2 units in our model potential energy changes the mean life by more than 30 orders of magnitude.

Step-by-step solution

Concept involved

Alpha decay is a radioactive decay in which an atom emits a helium nucleus also known as an alpha particle and hence it transforms into a different nucleus which differs in atomic number by two and mass number by 4 from its parent atom.

Given parameters

Kinetic energy given as,

$\begin{array}{rcl}E& =& 4.3MeV\\ & =& 6.8\times {10}^{-13}J\end{array}$

Potential Energy,

$\begin{array}{rcl}{U}_0& =& 17.5Mev\\ & =& 2.8\times {10}^{-12}J\end{array}$

Penetration depth,

$\begin{array}{rcl}L& =& \left(\frac{35}{4.3}-1\right)\times 7.4\times {10}^{-15}m\\ & =& 5.28\times {10}^{-14}m\end{array}$

Step 3(a): Determining life time

Here,

$\begin{array}{rcl}\sqrt{\frac{2m\left(U_0-E\right)}{\sout{h}}L}& =& \sqrt{\frac{2\left(4\times 1.66\times {10}^{-27}kg\right)\left(2.8\times {10}^{-12}J-6.8\times {10}^{-13}J\right)}{1.055\times {10}^{-34}Js}}5.28\times {10}^{-14}m\\ & =& 83.8\end{array}$

Hence, barrier is about 84 times the penetration depth.

Now, as you know that, the Tunneling probability is given by

$\begin{array}{rcl}T& =& 16\frac{E}{U_0}\left(1-\frac{E}{U_0}\right)e^{-2\sqrt{\frac{2m\left(U_0-E\right)}{\sout{h}}}L}\\ & =& 16\frac{4.3}{17.5}\left(1-\frac{4.3}{17.5}\right)e^{-2\times 84}\\ & =& 4.7\times {10}^{-52}\end{array}$

You also know that,

$\frac{numberofdecays}{time}=\frac{v}{2r_{nuc}}T$

Where, v = Velocity of the alpha particle

r_nuc= Radius of the nucleus

Now, if,

$E=\frac{1}{2}m{V}^2$

Where, m = mass of the alpha particle,

Velocity of the alpha particle can be calculated as,

$\begin{array}{rcl}& \Rightarrow & 6.88\times {10}^{-13}J=\frac{1}{2}\left(4\times 1.66\times {10}^{-27}\right)v^2\\ & \Rightarrow & v=1.44\times {10}^{7}m/s\end{array}$

Now,

$\begin{array}{rcl}\frac{v}{2r_{nuc}}T& =& \frac{1.44\times {10}^{7}m/s}{2\times 7.4\times {10}^{-15}m}4.7\times {10}^{-73}\\ & =& 4.6\times {10}^{-52}\end{array}$

Hence, at one decay every10⁵² seconds, the mean lifetime will be 10⁵² seconds or 10⁴⁴years.

Step 4(b): Determining value of U0

If the known value of the mean time to decay for uranium-238.

$\begin{array}{rcl}\frac{numberofdecays}{time}& =& \frac{1}{6.5\times {10}^{9}years\times 3.16\times {10}^{7}s/yr}\\ & =& 4.87\times {10}^{-18}{S}^{-1}\end{array}$

Now, if you set the above-mentioned value equal to:$\frac{v}{2r_{nuc}}T$, you get,

role="math" localid="1660046540630" $\begin{array}{rcl}4.87\times {10}^{-18}{s}^{-1}& =& 9.7\times {10}^{20}{s}^{-1}T\\ T& =& 5\times {10}^{-39}\\ T& \cong & e^{-2\sqrt{\frac{2m(u_0-E)}{\sout{h}}}L}\\ {10}^{-38}& =& \sqrt{\frac{2\left(4\times 1.66\times {10}^{-27}kg\right)\left(U_0-6.8*{10}^{-13}\right)}{1.055\times {10}^{-34}J.s}}5.28\times {10}^{-14}m\\ U_0& =& 1.3\times {10}^{-12}J\\ & =& 7.9MeV\end{array}$

Again, if you set the above-mentioned value equal to $16\frac{E}{U_0}\left(1-\frac{E}{U_0}\right)e^{-2\sqrt{\frac{2m(U_0-E)}{\sout{h}}}L}$, it is impossible to find the value of U₀, but the magnitude is significant and rest of the terms can be neglected. You can calculate U₀ as,

$\begin{array}{rcl}T& \cong & e^{-2\sqrt{\frac{2m\left(U_0-E\right)}{\sout{h}}}L}\\ {10}^{-38}& =& \sqrt{\frac{2\left(4\times 1.66\times {10}^{-27}kg\right)\left(U_0-6.8*{10}^{-13}\right)}{1.055\times {10}^{-34}J.s}}5.28\times {10}^{-14}m\\ U_0& =& 1.3\times {10}^{-12}J\\ & =& 7.9MeV\end{array}$

Hence,$U_0=1.3\times {10}^{-12}J=7.9MeV$

Step 4(c): Determining the sensitivity of potential barrier

It is observed in the previous steps that, the decay time is very sensitive with respect to the height of potential barrier.

A change of about 2 units in our model potential energy changes the mean life by more than 30 orders of magnitude.