A particle is confined between rigid walls separated by a distance L = 0.167 nm. The particle is in the second excited state (n = 3). Evaluate the probability to find the particle in an interval of width 1.00 pm located at: x = 0.166 nm; x = 0.028 nm; x = 0.067 nm. (Hint: No integrations are required for this problem; use P(x)dx = |ψ(x)|2dx directly.) What would be the corresponding results for a classical particle?

The answer for a classical particle is 0.00595

Explanation:

The equation of the wave function of a particle in a box in the second excited state equals:

ψ(x) = ((2/L)^1/2) * sin((3*pi*x)/L)

The probability is equal to:

P(x)dx = (|ψ(x)|^2)dx = ((2/L)^1/2) * sin((3*pi*x)/L) = (2/L) * sin^2((3*pi*x)/L) dx

for x = 0.166 nm

P(x)dx = (2/0.167) * sin^2((3*pi*0.166)/0.167) * 100 pm = 0.037x10^-3

for x = 0.028 nm

P(x)dx = (2/0.167) * sin^2((3*pi*0.028)/0.167) * 100 pm = 11x10^-3

for x = 0.067 nm

P(x)dx = (2/0.167) * sin^2((3*pi*0.067)/0.167) * 100 pm = 3.99x10^-3

therefore, the classical probability is equal to:

(1/L)dx = (1/0.167)*100 pm = 0.00595