Question

An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV. If the barrier width is 0.40 nm, what is the probability that the electron will tunnel through the barrier?

Answer 1

Probability = P(T) = 0.0142

Explanation:

It is given that the potential barrier is 4.70 eV and kinetic energy is 2.80 eV

and the barrier width is 0.40 nm

The probability of tunneling can be found using the equation below

P(T) = 16E/U₀( 1 - E/U₀)e^(-2βL)

β = √2m/h²(U₀ - KE)

Ehere m and h are constants with values m = 511 keV/c² and h = 0.1973 keV.nm/c

so

2m/h² = 2*(511)/(0.1973)²

2m/h² = 26.254 eV.nm²

β = √26.254 (4.70 - 2.80)

β = 7.06/nm

P(T) = 16E/U₀( 1 - E/U₀)e^(-2βL)

P(T) = 16*2.80/4.70( 1 - 2.80/4.70)e^(-2*7.06*0.40)

P(T) = 9.532*(0.404)e^(-5.6)

P(T) = 0.0142

Therefore, the probability that the electron will tunnel through the potential barrier of height 4.70 eV with kinetic energy of 2.80 eV and barrier width of 0.40 nm is 0.0142.