Question

If a proton of kinetic energy \(18.0 \mathrm{MeV}\) encounters a rectangular potential energy barrier of height \(29.8 \mathrm{MeV}\) and width \(1.00 \cdot 10^{-15} \mathrm{~m},\) what is the probability that the proton will tunnel through the barrier?

Short answer

The probability of the proton tunneling through the barrier is approximately 1.50 x 10^-208. This is an extremely small probability, indicating that the proton is very unlikely to tunnel through the barrier.

Step-by-step solution

Determine the Energy of the Proton Inside the Barrier

To find the energy of the proton inside the barrier, \(E_p\), we will use the equation: $$E_p = E_k - V,$$ where \(E_k = 18.0\,\text{MeV}\) and \(V = 29.8\,\text{MeV}\). Plugging in the values, we get: $$E_p = 18.0\,\text{MeV} - 29.8\,\text{MeV} = -11.8\,\text{MeV}.$$

Calculate the Wave Number, K

Now that we have the energy of the proton inside the barrier, we can use the equation for the wave number \(K\): $$K = \frac{\sqrt{2mE_p}}{\hbar},$$ where \(m\) is the mass of the proton (about \(1.67\times10^{-27}\text{kg}\)), \(E_p\) is the energy of the proton inside the barrier (-11.8 MeV or \(-1.89\times10^{-12}\text{J}\)), and \(\hbar\) is the reduced Planck constant (about \(1.054\times10^{-34}\text{J}\cdot\text{s}\)). Plugging in the values, we get: $$K = \frac{\sqrt{2(1.67\times10^{-27}\text{kg})(-1.89\times10^{-12}\text{J})}}{1.054\times10^{-34}\text{J}\cdot\text{s}}.$$ Since the energy is negative, we need to use its absolute value: $$K = \frac{\sqrt{2(1.67\times10^{-27}\text{kg})(1.89\times10^{-12}\text{J})}}{1.054\times10^{-34}\text{J}\cdot\text{s}} \approx 2.39\times10^{13}\,m^{-1}.$$

Calculate the Transmission Coefficient

Now that we have the wave number \(K\), we can use the equation for the transmission coefficient \(T\): $$T = e^{-2Kd},$$ where \(d = 1.00\times 10^{-15}\,\text{m}\). Plugging in the values, we find the tunneling probability: $$T = e^{-2(2.39\times10^{13}\,m^{-1})(1.00\times10^{-15}\,\text{m})} \approx 1.50\times10^{-208}.$$ Therefore, the probability that the proton will tunnel through the barrier is approximately \(1.50\times10^{-208}\). This is an extremely small probability, indicating that the proton is very unlikely to tunnel through the barrier.