Question

The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose the box is 100 nm long.Calculate the probability that the particle is (a) between x= 4.95 nm and 5.05 nm (b) between x= 1.95 nm and 2.05 nm (c) between x=9.90 nm and 10.00 nm (d) in the right half of the box (e) in the central third of the box.

Answer 1

To find the probability that the particle is between two positions, we need to integrate the probability density function over that region. We can use the given wave function to calculate the probabilities for different intervals.

Step-by-step explanation:

To calculate the probability that the particle is between two positions, we need to find the integral of the probability density function (|Ψ|^2) over that region.

a) To find the probability between x = 4.95 nm and 5.05 nm, we need to integrate |Ψ|^2 from x = 4.95 nm to x = 5.05 nm.

b) To find the probability between x = 1.95 nm and 2.05 nm, we need to integrate |Ψ|^2 from x = 1.95 nm to x = 2.05 nm.

c) To find the probability between x = 9.90 nm and 10.00 nm, we need to integrate |Ψ|^2 from x = 9.90 nm to x = 10.00 nm.

d) To find the probability in the right half of the box, we need to integrate |Ψ|^2 from x = L/2 to x = L.

e) To find the probability in the central third of the box, we need to integrate |Ψ|^2 from x = L/3 to x = 2L/3.

Answer 2

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²

Step-by-step explanation:

The probability (P) to find the particle is given by:

`P=\int_{x_(1)}^{x_(2)}(\Psi\cdot \Psi) dx = \int_{x_(1)}^{x_(2)} ((2/L)^(1/2) Sin(\pi x/L))^(2)dx`

`P = \int_{x_(1)}^{x_(2)} (2/L) Sin^(2)(\pi x/L)dx` (1)

The solution of the intregral of equation (1) is:

`P=(2)/(L) [(X)/(2) - (Sin(2\pi x/L))/(4\pi /L)]|_{x_(1)}^{x_(2)}`

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

`P=(2)/(100) [(X)/(2) - (Sin(2\pi x/100))/(4\pi /100)]|_(4.95)^(5.05) = 4.98 \cdot 10^(-5)`

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

`P=(2)/(100) [(X)/(2) - (Sin(2\pi x/100))/(4\pi /100)]|_(1.95)^(2.05) = 7.89 \cdot 10^(-6)`

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

`P=(2)/(100) [(X)/(2) - (Sin(2\pi x/100))/(4\pi /100)]|_(9.90)^(10.00) = 1.89 \cdot 10^(-4)`

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

`P=(2)/(100) [(X)/(2) - (Sin(2\pi x/100))/(4\pi /100)]|_(0)^(50.00) = 0.5`

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

`P=(2)/(100) [(X)/(2) - (Sin(2\pi x/100))/(4\pi /100)]|_(0)^(16.7) = 2.9 \cdot 10^(-2)`

I hope it helps you!