Question

A 8.59-MeV alpha particle (mass \(=3.7274 \mathrm{GeV} / \mathrm{c}^{2}\) ) inside a heavy nucleus encounters a barrier whose average height is \(15.9 \mathrm{MeV}\). The tunneling probability is measured to be \(1.042 \cdot 10^{-18} .\) What is the width of the barrier? (Hint: A potentially useful value is \(\hbar c=197.327 \mathrm{MeV} \mathrm{fm} .)\)

Short answer

Answer: The width of the potential barrier is approximately 57.12 fm (femtometer).

Step-by-step solution

Known and unknown values

First, let's list down all the given values: - Energy of the alpha particle (E): 8.59 MeV - Mass of the alpha particle (m): 3.7274 GeV/c^2 - Height of the barrier (V_0): 15.9 MeV - Tunneling probability (T): 1.042 × 10^(-18) Additionally, we have the value of ħc (Planck's constant, h, divided by 2π, times the speed of light): 197.327 MeV·fm. We need to find the width of the barrier (d).

Determine k and K values

We will use the following equations to determine the values of k and K representing the wave numbers of regions inside and outside the potential barrier, respectively: k = sqrt(2mE/ħ^2) K = sqrt(2m(V_0 - E)/ħ^2)

Conversion of units

Before proceeding with calculations, let's convert the mass and energies into their respective units: E = 8.59 MeV × (1 GeV / 1000 MeV) = 8.59 × 10^(-3) GeV m = 3.7274 GeV/c^2 Now, let's compute k and K: k = sqrt(2 × 3.7274 × 8.59 × 10^(-3) / (197.327 fm)^2) k ≈ 0.053823 fm^(-1) K = sqrt(2 × 3.7274 × (15.9 - 8.59) × 10^(-3) / (197.327 fm)^2) K ≈ 0.079496 fm^(-1)

Calculate the width of the barrier

Now, we will use the tunneling probability (T) and the wave number values (k and K) to determine the width of the barrier (d): T = e^(-2Kd) From this, we can identify d: d = (1/(-2K)) × ln(T) d ≈ (1/(-2 × 0.079496)) × ln(1.042 × 10^(-18)) d ≈ 57.12 fm Therefore, the width of the barrier is approximately 57.12 fm (femtometer).