Question

Four quantum particles, each with energy E, approach the potential-energy barriers seen in FIGURE Q40.8 from the left. Rank in order, from largest to smallest, the tunneling probabilities${\left(P_{\text{tunnel}}\right)}_{\mathrm{a}}\text{to}{\left(P_{\text{tunnel}}\right)}_{\mathrm{d}}$

Short answer

The rank of tunneling probabilities from largest value to smallest value,

$P_{\text{tunnel}\left(\mathrm{d}\right)}>P_{\text{tunnel (a)}}>P_{\text{tunnel (b)}}>\text{and}{P}_{\text{tunnel}\left(\mathrm{c}\right)}$

Step-by-step solution

Step 1. Given information

Equation for penetration distance=

$\eta =\frac{\hslash }{\sqrt{2m\left(U_0-E\right)}}$

Here,

$\hslash$= reduced Planck's constant,

$m$= mass of the particle,

$U_0$=height of the potential barrier, and

$E=$energy of the incident particle.

Probability of tunneling through the potential barrier,

$P_{\text{tunnel}}=e^{-\frac{w}{\eta }}$

Substituting $\frac{\hslash }{\sqrt{2m\left(U_0-E\right)}}\text{=}\eta$,

$P_{\text{tunnel}}=e^{\frac{2w}{\sqrt{2m\left(U_0-E\right)}}}$

$=e^{\frac{2\sqrt{2m\left(U_0-E\right)}}{\hslash }w}$

Step 2. Finding the probability of tunneling through barrier (a)

$P_{\text{tunnel (a)}}=e^{\frac{2\sqrt{2m\left(U_0-E\right)}}{\hslash }w}$

$P_{\text{tunnel (a)}}=e^{\frac{2\sqrt{2\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(1\mathrm{eV}\right)\left(\frac{1.60\times {10}^{-19}\mathrm{J}}{1\mathrm{cV}}\right)}\left({10}^{-9}\mathrm{m}\right)}{1.05\times {10}^{-34}\mathrm{Js}}}$

$=3.4165\times {10}^{-5}$

Step 3. Finding the probability of tunneling through barrier (b)

$P_{\text{tunnel (b)}}=e^{\frac{2\sqrt{2m\left(U_0-E\right)}}{h}w}$

$P_{\text{tunnel (b)}}=e^{\frac{2\sqrt{2\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(2\mathrm{eV}\right)\left(\frac{1.60\times {10}^{-19}\mathrm{J}}{1\mathrm{cV}}\right)}}{1.05\times {10}^{-34}\mathrm{J}\mathrm{s}}\left({10}^{-9}\mathrm{m}\right)}$

$=4.815\times {10}^{-7}$

Step 4. Finding the probability of tunneling through barrier (c)

$P_{\text{tunnel (c)}}=e^{\frac{2\sqrt{2m\left(U_0-E\right)}}{h}w}$

$P_{\text{tunnel (c)}}=e^{\frac{2\sqrt{2\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(1\mathrm{eV}\right)\left(\frac{1.60\times {10}^{-19}\mathrm{J}}{1\mathrm{cV}}\right)}}{1.05\times {10}^{-34}\mathrm{J}-\mathrm{s}}\left(2\times {10}^{-9}\mathrm{m}\right)}$

$=1.167\times {10}^{-9}$

Step 5. Finding the probability of tunneling through barrier (d)

$P_{\text{tunnel (d)}}=e^{\frac{2\sqrt{2m\left(U_0-E\right)}}{\hslash }w}$

$P_{\text{tunnel (d)}}=e^{\frac{2\sqrt{2\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(2\mathrm{eV}\right)\left(\frac{1.60\mathrm{b}{10}^{-19}\mathrm{J}}{1\mathrm{eV}}\right)}}{1.05\times {10}^{-34}\mathrm{J}·\mathrm{s}}\left(0.5\times {10}^{-9}\mathrm{m}\right)}$

$=6.9465\times {10}^{-4}$

Thus, the rank of probabilities of tunneling through the barriers from largest to smallest,

$P_{\text{tunnel}\left(\mathrm{d}\right)}>P_{\text{tunnel (a)}}>P_{\text{tunnel}\left(\mathrm{b}\right)}>\text{and}{P}_{\text{tunnel}\left(\mathrm{c}\right)}$