Question

Neutrons of energy 8 MeV are incident on a potential step of height 48 MeV. As they penetrate the classically forbidden region, the distance at which the probability density of finding neutrons decreases by a factor of 100 is \rule{1cm{0.15mm} fm. (up to two decimal places)
(Take \(\hbar c = 200\) MeV fm, and the rest mass energy of neutron = 1 GeV.)}

Hint:

In quantum tunneling problems, the probability density \(|\psi|^2\) decays as \(e^{-2\kappa x}\). A common mistake is to forget the factor of 2. Remember, the wavefunction \(\psi\) decays as \(e^{-\kappa x}\), so the probability density decays twice as fast. Using quantities like \(\hbar c\) and \(mc^2\) simplifies calculations and avoids unit conversion issues.

Answer 1

Correct Answer: 1.63

Step 1: Understanding the Concept:
This problem describes quantum tunneling into a potential barrier. When a particle with energy \(E\) encounters a potential barrier \(V_0\) where \(E < V_0\), its wavefunction does not drop to zero immediately at the boundary but decays exponentially into the barrier. The probability density, which is the square of the wavefunction's magnitude, also decays exponentially. 
Step 2: Key Formula or Approach:  
In the classically forbidden region (\(x > 0\)), the wavefunction \(\psi(x)\) has the form: \[ \psi(x) = A e^{-\kappa x} \] The probability density is \(P(x) = |\psi(x)|^2 = |A|^2 e^{-2\kappa x}\). The decay constant \(\kappa\) is given by: \[ \kappa = \frac{\sqrt{2m(V_0 - E)}}{\hbar} \] where \(m\) is the mass of the particle. It is convenient to use relativistic units by rewriting \(\kappa\) as: \[ \kappa = \frac{\sqrt{2(mc^2)(V_0 - E)}}{\hbar c} \] Step 3: Detailed Explanation: 
Given values: Particle energy, \(E = 8\) MeV. Potential height, \(V_0 = 48\) MeV. Neutron rest mass energy, \(mc^2 = 1\) GeV = 1000 MeV. \(\hbar c = 200\) MeV·fm. First, calculate the decay constant \(\kappa\): \[ V_0 - E = 48 \text{ MeV} - 8 \text{ MeV} = 40 \text{ MeV} \] \[ \kappa = \frac{\sqrt{2 \times (1000 \text{ MeV}) \times (40 \text{ MeV})}}{200 \text{ MeV fm}} = \frac{\sqrt{80000}}{200} \text{ fm}^{-1} \] \[ \kappa = \frac{\sqrt{16 \times 5000}}{200} = \frac{4 \sqrt{5000}}{200} = \frac{\sqrt{5000}}{50} = \frac{\sqrt{2500 \times 2}}{50} = \frac{50\sqrt{2}}{50} = \sqrt{2} \text{ fm}^{-1} \] So, \(\kappa \approx 1.414\) fm\(^{-1}\). The probability density \(P(x)\) decreases by a factor of 100 relative to its value at the boundary \(P(0)\): \[ \frac{P(x)}{P(0)} = \frac{|A|^2 e^{-2\kappa x}}{|A|^2 e^{0}} = e^{-2\kappa x} = \frac{1}{100} \] To find the distance \(x\), we solve for \(x\): \[ -2\kappa x = \ln\left(\frac{1}{100}\right) = -\ln(100) \] \[ x = \frac{\ln(100)}{2\kappa} \] Substitute the value of \(\kappa\): \[ x = \frac{\ln(100)}{2\sqrt{2}} \text{ fm} \] Using \(\ln(100) \approx 4.60517\) and \(\sqrt{2} \approx 1.41421\): \[ x \approx \frac{4.60517}{2 \times 1.41421} = \frac{4.60517}{2.82842} \approx 1.628 \text{ fm} \] Step 4: Final Answer: 
The distance at which the probability density decreases by a factor of 100 is 1.63 fm.