Answer to Question #121947 in Quantum Mechanics for Ranil

Question

A proton and an alpha particle attempt to penetrate a rectangular potential barrier of height
10 MeV and thickness 10-14m. Both particles have total energies of 5MeV. Use qualitative
arguments to predict which particle has the highest probability of succeeding and evaluate
quantitatively the probability of success of both particles.

Answer 1

As, both particles have equal energies and $K = \frac{p^ 2}{2m}$ , thus. the momentum of the alpha particle (since, more massive particle than proton) will be smaller, and the velocity of the proton will be greater. More is the velocity ,thus proton is more likely to get penetrate the barrier.

Now, we know that,

$$k_{p}a = \frac{\sqrt{2m_p(V_0 − E)}}{\hbar}a\\ k_{\alpha} a=\frac{\sqrt{2m_{\alpha}(V_0 − E)}}{\hbar}a$$

Thus,

$$k_pa= \frac{\sqrt{2(938.3 MeV/c^22)(5 MeV)}} {0.6582 × 10^{−15} eV · s} (10^{−14}m)=4.9089$$

Since, mass of $\alpha$ particle is $m_{\alpha}\approx4m_p$ , thus

$$k_{\alpha}a=2k_pa=9.8179$$

Now, we have to calculate the transmission coefficient for both particles to calculate the probability,Hence

$$T=\bigg(1+\frac{\sinh^2(ka)}{4\frac{E}{V_0}(1-\frac{E}{V_0})}\bigg)^{-1}$$

Thus, on plugin the values , we get

$$T_p=2.17869438\times 10^{-4}\\ T_{\alpha}=1.1866985\times 10^{-8}$$

Thus, corresponding probability is

$$0.00217\%\\ 0.000000118\%$$

respectively.