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

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.