Question

is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, m el = 9.11 × 10-31 kg, ℏℏ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s) 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? (1 eV = 1.60 × 10-19 J, m el = 9.11 × 10-31 kg, = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s) 3.5 × 10-3 7.3 × 10-3 1.4 × 10-2 3.5 × 10-2 2.9 × 10-3

Answer 1

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?

a) 3.5 × 10-3

b) 7.3 × 10-3

c) 1.4 × 10-2

d) 3.5 × 10-2

e) 2.9 × 10-3

Given Information:

Potential barrier = U₀ = 4.70 eV

kinetic energy = E = 2.80 eV

Barrier width = L = 0.40 nm

Required Information:

Probability of tunneling = ?

Answer:

c) Probability of tunneling = 1.42×10⁻²

Explanation:

The probability of tunneling is given by

T(L,E) = 16E/U₀( 1 - E/U₀)e^(-2βL)

Where β is given by

β = √2m/h²(U₀ - KE)

Where m is the mass of electron and h is Planck’s constant

m = 511 keV/c² and h = 0.1973 keV.nm/c

2m/h² = 2*(511)/(0.1973)²

2m/h² = 26.254 eV.nm²

β = √26.254 (4.70 - 2.80)

β = 7.06/nm

T(L,E) = 16E/U₀( 1 - E/U₀)e^(-2βL)

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

T(L,E) = 9.532*(0.404)e^(-5.6)

T(L,E) = 0.0142

or

T(L,E) = 1.42×10⁻²