A 6.0-eV electron impacts on a barrier with height \(11.0 \mathrm{eV} .\) Find the probability of the electron to tunnel through the barrier if the barrier width is (a) \(0.80 \mathrm{nm}\) and (b) \(0.40 \mathrm{nm}\)

First, we must determine the wave vector inside the barrier, using the formula: \[k_2 = \sqrt{\frac{2m(V - E)}{\hbar^2}}\] where m is the mass of the electron, which is approximately \(9.11 \times 10^{-31} kg\), and \(\hbar\), the reduced Planck constant, is approximately \(1.055 \times 10^{-34} J⋅s\). We are given E and V, so we can plug in the values: \[k_2 = \sqrt{\frac{2(9.11 \times 10^{-31} kg)(11.0 eV - 6.0 eV)}{(1.055 \times 10^{-34} J⋅s)^2}}\] Note that we must convert eV to J in order to use the formula correctly: \[1 eV = 1.602 \times 10^{-19} J\] Therefore, \[k_2 = \sqrt{\frac{2(9.11 \times 10^{-31} kg)(5.0 \times 1.602 \times 10^{-19} J)}{(1.055 \times 10^{-34} J⋅s)^2}} = 1.241 \times 10^{10} m^{-1}\] Now we have the wave vector, k_2.

Next, we calculate the transmission probability using the given formula when the barrier width, d, is 0.80 nm: \[T(E) = \frac{1}{1 + \frac{V^2 \sinh^2(k_2 d)}{4E(V - E)}}\] Plugging in the values and converting d to meters: \[T(E) = \frac{1}{1 + \frac{(11.0 \times 1.602 \times 10^{-19} J)^2 \sinh^2(1.241 \times 10^{10} m^{-1} \times 0.8 \times 10^{-9} m)}{4(6.0 \times 1.602 \times 10^{-19} J)(5.0 \times 1.602 \times 10^{-19} J)}}\] \[T(E) ≈ 3.90 \times 10^{-6}\] Thus, the probability of the electron tunneling through the barrier with a width of 0.80 nm is approximately \(3.90 \times 10^{-6}\).

Finally, we calculate the transmission probability when the barrier width, d, is 0.40 nm: \[T(E) = \frac{1}{1 + \frac{V^2 \sinh^2(k_2 d)}{4E(V - E)}}\] Plugging in the values and converting d to meters: \[T(E) = \frac{1}{1 + \frac{(11.0 \times 1.602 \times 10^{-19} J)^2 \sinh^2(1.241 \times 10^{10} m^{-1} \times 0.4 \times 10^{-9} m)}{4(6.0 \times 1.602 \times 10^{-19} J)(5.0 \times 1.602 \times 10^{-19} J)}}\] \[T(E) ≈ 2.58 \times 10^{-3}\] Thus, the probability of the electron tunneling through the barrier with a width of 0.40 nm is approximately \(2.58 \times 10^{-3}\).