AP Physics 2 Exam Question - Quantum Phenomena

A particle of mass m is trapped in an infinite potential well of width L, which can be visualized as a one-dimensional box. The particle is initially in its ground state with a wave function given by:

ψ(x) = √(2/L) * sin(πx/L)

where x is the position within the well.

a) Calculate the probability of finding the particle between x = L/4 and x = L/2.

b) What is the expectation value of the position of the particle in the ground state?

c) Determine the energy of the excited state corresponding to the first excited state of the particle in the one-dimensional box.

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Answer:

a) To find the probability of finding the particle between x = L/4 and x = L/2, we need to integrate the wave function squared over this range and normalize it with respect to the entire well. The probability P is given by:

P = ∫(L/4 to L/2) |ψ(x)|^2 dx

Substituting the given wave function, we have:

P = ∫(L/4 to L/2) |√(2/L) * sin(πx/L)|^2 dx

Simplifying, we get:

P = ∫(L/4 to L/2) (2/L) * sin^2(πx/L) dx

Using the identity sin^2θ = (1/2)(1 - cos2θ), the probability becomes:

P = ∫(L/4 to L/2) (1/L) * (1 - cos(2πx/L)) dx

Evaluating the integral, we have:

P = (1/L) * ∫(L/4 to L/2) dx - (1/L) * ∫(L/4 to L/2) cos(2πx/L) dx

P = (1/L) * [(x/L) from L/4 to L/2] - (1/L) * [1/(2π/L) * sin(2πx/L)] from L/4 to L/2

P = (1/L) * [(L/2 - L/4) - (sin(π) - sin(π/2))]

P = (1/L) * (L/4 + 0)

P = 1/4

Therefore, the probability of finding the particle between x = L/4 and x = L/2 is 1/4.

b) The expectation value of the position of the particle in the ground state can be found using the formula:

< x > = ∫ x |ψ|^2 dx

Substituting the given wave function, we have:

< x > = ∫ x * (√(2/L) * sin(πx/L))^2 dx

Simplifying, we get:

< x > = ∫ (2/L) * x * sin^2(πx/L) dx

Using the identity sin^2θ = (1/2)(1 - cos2θ), the expectation value becomes:

< x > = ∫ (1/L) * x * (1 - cos(2πx/L)) dx

Splitting the integral into two parts, we evaluate each term:

Term 1: ∫ (1/L) * x dx = (1/L) * (1/2)x^2 = x^2 / 2L

Term 2: - ∫ (1/2L) * x * cos(2πx/L) dx

To integrate the second term, we use integration by parts with u = x and dv = (1/2L) * cos(2πx/L) dx:

du = dx, v = (1/2π) * sin(2πx/L)

Applying the integration by parts formula, we have:

• ∫ (1/2L) * x * cos(2πx/L) dx = - [(1/2L) * x * (1/2π) * sin(2πx/L)] + ∫ (1/2π) * sin(2πx/L) dx

Simplifying, we get:

• ∫ (1/2L) * x * cos(2πx/L) dx = - [(1/4Lπ) * x * sin(2πx/L)] + (-1/(4π)) * (L/2π) * cos(2πx/L)

Evaluating the integral, we have:

• ∫ (1/2L) * x * cos(2πx/L) dx = - [(1/4Lπ) * x * sin(2πx/L)] + (-1/(4π)) * (L/2π) * cos(2πx/L) + C

Adding both terms:

< x > = (x^2 / 2L) - [(1/4Lπ) * x * sin(2πx/L)] + (-1/(4π)) * (L/2π) * cos(2πx/L) + C

To find the constant C, we can apply the normalization condition. Since the wave function is normalized, the integral of |ψ|^2 over the entire well L should be equal to 1:

∫ |ψ|^2 dx = 1

∫ (√(2/L) * sin(πx/L))^2 dx = 1

∫ (2/L) * sin^2(πx/L) dx = 1

Using the sine identity, we have:

∫ (1/L) * (1 - cos(2πx/L)) dx = 1

Integrating, we get:

(1/L) [(x - (L/2π) * sin(2πx/L))] from 0 to L = 1

Simplifying, we have:

(1/L) [(L - (L/2π) * sin(2π)) - (0 - (L/2π) * sin(0))] = 1

(1/L) [(L + (L/2π) * sin(0))] = 1

(1/L) [(L + 0)] = 1

(L/L) = 1

1 = 1

Therefore, the expectation value equation becomes:

< x > = (x^2 / 2L) - [(1/4Lπ) * x * sin(2πx/L)] + (-1/(4π)) * (L/2π) * cos(2πx/L)

Now, we evaluate < x > from 0 to L using this equation:

< x > = [(L^2 / 2L) - (0^2 / 2L)] - [(1/4Lπ) * L * sin(2πL/L) - (1/4Lπ) * 0 * sin(2π(0)/L)] + (-1/(4π)) * (L/2π) * cos(2πL/L) - (-1/(4π)) * (L/2π) * cos(2π(0)/L)

< x > = (L/2) - (1/4Lπ) * L * sin(2π) + (-1/(4π)) * (L/2π) * cos(0) - 0

Since sin(2π) = sin(0) = 0 and cos(0) = 1, we have:

< x > = (L/2) + 0 - (1/(4π)) * (L/2π)

< x > = (L/2) - (1/8π^2) * L

Therefore, the expectation value of the position of the particle in the ground state is (L/2) - (1/8π^2) * L.

c) The energy of the excited state corresponding to the first excited state of the particle in the one-dimensional box can be determined using the formula:

E₁ = (n²h²)/(8mL²)

where n is the quantum number (1 for first excited state), h is Planck's constant, m is the mass of the particle, and L is the width of the potential well.

Substituting the provided values and solving, we have:

E₁ = (1²h²)/(8mL²)

E₁ = (h²)/(8mL²)

Therefore, the energy of the first excited state is (h²)/(8mL²).

This completes the solution for the given AP Physics 2 Exam question on Quantum Phenomena.