Part 1: Finding the Normalisation Constant A The wave function must be normalized, which means that the integral of the absolute square of the wave function over all space must be equal to 1. In mathematical terms, this is expressed as: ∫|ψ(x)|² dx = 1 For the given wave function ψ(x) = Ax, we square it to get |ψ(x)|² = A²x². We then integrate this from 0 to 1 (the limits given in the problem): ∫ (from 0 to 1) A²x² dx = 1 This integral evaluates to (A²/3)[x³] (from 0 to 1) = A²/3. Setting this equal to 1, we solve for A: A²/3 = 1 A² = 3 A = sqrt(3) So, the normalization constant A is sqrt(3). Part 2: Finding the Expectation Value of the Particle's Position The expectation value of the particle's position is given by the integral of x times the absolute square of the wave function over all space. In mathematical terms, this is expressed as: = ∫ x|ψ(x)|² dx For the given wave function ψ(x) = Ax (with A = sqrt(3)), we have |ψ(x)|² = 3x². So, we compute the integral: = ∫ (from 0 to 1) x(3x²) dx = ∫ (from 0 to 1) 3x³ dx This integral evaluates to (3/4)[x⁴] (from 0 to 1) = 3/4. So, the expectation value of the particle's position is 3/4. Part 3: Finding the Probability that the Particle will be Found between x = 0.245 and x = 0.465 The probability that the particle will be found in a certain interval [a, b] is given by the integral of the absolute square of the wave function over that interval. In mathematical terms, this is expressed as: P(a ≤ x ≤ b) = ∫ (from a to b) |ψ(x)|² dx For the given wave function ψ(x) = Ax (with A = sqrt(3)), we have |ψ(x)|² = 3x². So, we compute the integral: P(0.245 ≤ x ≤ 0.465) = ∫ (from 0.245 to 0.465) 3x² dx This integral evaluates to (x³) (from 0.245 to 0.465) = 0.465³ - 0.245³. So, the probability that the particle will be found between x = 0.245 and x = 0.465 is 0.465³ - 0.245³.

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Part 1: Finding the Normalisation Constant A The wave function must be normalized, which means that the integral of the absolute square of the wave function over all space must be equal to 1. In mathematical terms, this is expressed as: ∫|ψ(x)|² dx = 1 For the given wave function ψ(x) = Ax, we square it to get |ψ(x)|² = A²x². We then integrate this from 0 to 1 (the limits given in the problem): ∫ (from 0 to 1) A²x² dx = 1 This integral evaluates to (A²/3)[x³] (from 0 to 1) = A²/3. Setting this equal to 1, we solve for A: A²/3 = 1 A² = 3 A = sqrt(3) So, the normalization constant A is sqrt(3). Part 2: Finding the Expectation Value of the Particle's Position The expectation value of the particle's position is given by the integral of x times the absolute square of the wave function over all space. In mathematical terms, this is expressed as: = ∫ x|ψ(x)|² dx For the given wave function ψ(x) = Ax (with A = sqrt(3)), we have |ψ(x)|² = 3x². So, we compute the integral: = ∫ (from 0 to 1) x(3x²) dx = ∫ (from 0 to 1) 3x³ dx This integral evaluates to (3/4)[x⁴] (from 0 to 1) = 3/4. So, the expectation value of the particle's position is 3/4. Part 3: Finding the Probability that the Particle will be Found between x = 0.245 and x = 0.465 The probability that the particle will be found in a certain interval [a, b] is given by the integral of the absolute square of the wave function over that interval. In mathematical terms, this is expressed as: P(a ≤ x ≤ b) = ∫ (from a to b) |ψ(x)|² dx For the given wave function ψ(x) = Ax (with A = sqrt(3)), we have |ψ(x)|² = 3x². So, we compute the integral: P(0.245 ≤ x ≤ 0.465) = ∫ (from 0.245 to 0.465) 3x² dx This integral evaluates to (x³) (from 0.245 to 0.465) = 0.465³ - 0.245³. So, the probability that the particle will be found between x = 0.245 and x = 0.465 is 0.465³ - 0.245³.

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