Question

Suppose a particle of mass m is confined to a one-dimensional box of length L. We can model this as an infinite square well in which the particle's potential energy inside the box is zero and the potential energy outside is infinite. For a particle in its first excited state, what is the probability Prob(center20%) of finding the particle within the center 20% of the box

Answer 1

Answer: P = 4.86 × 10⁻²

Therefore, the particle's quantum number is 4.86 × 10⁻²

Step-by-step explanation:

The expression of wave function for a particle in one dimensional box is given as;

φ(x) = ( √2/L ) sin ( nπx/L )

now we input our given figures, the limit of the particle to find it within the center of the box is

xₓ = L/2 + 20% of L/2

xₓ = L/2 + (0.2)L/2

xₓ = 3L/5

And the lower limit is,

x₁ = L/2 - 20% of L/2

x₁ = L/2 - (0.2) L/2

x₁ = 2L / 5

The expression for the probability of finding the particle within the center of the box is

P = ∫ˣˣₓ₁ ║φ(x)║² dx

P = ∫ ³L/⁵ ₂L/₅║(√2/L) sin ( nπx/L)║²dx

= 2/L ( ∫ ³L/⁵ ₂L/₅║sin ( nπx/L)║²dx

= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2πnx/L)/2) dx)

The particle is in its first excited state, then

n =2

Then calculate the particle's quantum number as follows;

= 2/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 2π(2)x/L)/(2)) dx)

= 1/L ( ∫ ³L/⁵ ₂L/₅ (( 1 - cos ( 4πx/L)/2) dx)

= 1/L ( x - (L/4π)sin (4πx/L)) ³L/⁵ ₂L/₅

= 1/L ((3L/5) - (L/4π) sin (( 4π(3L/5)/L)) - (( 2L/5) - (L/4π)sin ( 4π(2L/5)/L)))

= 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))

Use the trigonometric formula to solve the above equation

sinA - sinB = 2sin ( A-B/2) cos (A+B/2)

Calculate the particle's quantum number as follows

P = 1/L ( L/5 + L/4π (sin(8π/5) - sin ( 12π/5)))

= 1/5 + 1/4π ( 2sin( (8π/5 -12π/5 ) / 2 ) cos ( (8π/5 + 12π/5) / 2 ))

= 1/5 + 1/2π ( -sin(2π/5) cos2π

= 1/5 - 1/2π ( sin (2π/5)(1))

= 0.0486 (10⁻²)(10²)

= 4.86 × 10⁻²

Therefore, the particle's quantum number is 4.86 × 10⁻²