Calculate the probability that a particle in a one-dimensional box of length aa is found between 0.29a0.29a and 0.34a0.34a when it is described by the following wave function: 2a−−√sin(πxa)2asin⁡(πxa). Express your answer to two significant figures.

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.

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic

i think the answer is true

step-by-step explanation: