Answer:
P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
Assuming the box is of unit length, a = 1,
0.29a = 0.29 and 0.34a = 0.34
P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = 0.167 = 0.17 to 2 s.f
Step-by-step explanation:
ψ(x) = 2a(√sin(πxa)
The probability of finding a particle in a specific position for a given energy level in a one-dimensional box is related to the square of the wavefunction
The probability of finding a particle between two points 0.29a and 0.34a is given mathematically as
P(0.29a < X < 0.34a) = ∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx
That is, integrating from 0.29a to 0.34a
ψ²(x) = [2a(√sin(πxa)]² = 4a² sin(πxa)
∫⁰•³⁴ᵃ₀.₂₉ₐ ψ²(x) dx = ∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa)) dx
∫⁰•³⁴ᵃ₀.₂₉ₐ (4a² sin(πxa))
= - [(4a²/πa)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ
- [(4a/π)cos(πxa)]⁰•³⁴ᵃ₀.₂₉ₐ = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
P(0.29a < X < 0.34a) = -(4a/π) [(cos 0.34πa²) - (cos 0.29πa²)]
Assuming the box is of unit length, a = 1
P(0.29 < X < 0.34) = (-4/π) [(cos 0.34π) - (cos 0.29π)] = (-1.273) [0.482 - 0.613] = 0.167.
