Question

Consider an electron approaching a potential barrier \(2.00 \mathrm{nm}\) wide and \(7.00 \mathrm{eV}\) high. What is the energy of the electron if it has a \(10.0 \%\) probability of tunneling through this barrier?

Short answer

E ≈ 2.46 eV #Conclusion# The energy of the electron with a 10.0% probability of tunneling through the potential barrier is approximately 2.46 eV.

Step-by-step solution

Write down the tunneling probability formula

The tunneling probability, T, of an electron with an energy E through a potential barrier of width L and height V is given by the following formula: T = \(e^{-2kL}\) where \(k=\sqrt{({2m(V-E)}/{\hbar^2})}\) m is the mass of the electron, and \(\hbar\) is the reduced Planck's constant. For the given problem, T = 10% = 0.1, L = 2.00 nm, and V = 7.00 eV. We need to find the energy of the electron, E.

Convert the given units to SI units

In order to work with SI units, we need to convert the given values. Potential barrier height, V = 7.00 eV = 7.00 * 1.6 * 10^{-19} J Barrier width, L = 2.00 nm = 2.00 * 10^{-9} m

Equation rearrangement and substitution

We need to find the value of E. Rearrange and substitute in the given values: \(0.1 = e^{-2*\sqrt{((2*9.11*10^{-31}*(7 * 1.6 * 10^{-19} - E))/(6.626*10^{-34}/(2*\pi))^2}) * 2 * 10^{-9}}\) Since we're dealing with natural exponentials, take the natural log of both sides: \(-2.303 = -2kL = -2*\sqrt{((2*9.11*10^{-31}*(7 * 1.6 * 10^{-19} - E))/(6.626*10^{-34}/(2*\pi))^2}) * 2 * 10^{-9}\) Now we can solve for E by squaring both sides and isolating E in the equation.

Solve for energy E

After isolating E and simplifying the equation, we have: E = \(7 * 1.6 * 10^{-19} - (99.248 * (6.626 * 10^{-34}/(2*\pi))^2)/(2 * 2 * 9.11 * 10^{-31} * 10^{-18})\) Now calculate the electron energy, E: