Probability Density in an infinite 1D square well

[gv3]

Homework Statement

The wave function of a particle of mass m confined in an infinite one-dimensional square well of width L = 0.23 nm, is:
ψ(x) = (2/L)1/2 sin(3πx/L) for 0 < x < L

ψ(x) = 0 everywhere else. The energy of the particle in this state is E = 63.974 eV.

1) What is the rest energy (mc²) of the particle?
ANS: 511133.6 ev

2) P(x) is the probability density.
That is, P(x)dx is the probability that the particle is between x and x+dx when dx is small.
For how many values of x does P(x) = 5 nm-1?
ANS: 6 values of x

3)What is the largest value of x for which P(x) = 5 nm-1?
ANS: ?

Homework Equations

P(x)=∫|ψ(x)|²dx

E=h²n²/(8mL²)

The Attempt at a Solution

For question 2 i only know its 6 because the question was multiple attempts. i don't know how to get that answer mathematically. I thought that by integrating P(x) squared and then solving for the x's would give me the 6 values of x but its not making sense to me. After i have the 6 values, then the answer to question 3 is just the largest value of x.

 

[kuruman]

For how many values of x does P(x) = 5 nm-1?

You have all you need.
P(x)dx= ψ^*ψ dx = (2/L) sin²(3πx/L)dx
Solve
P(x) = 5 nm^-1

 

[gv3]

You have all you need.
P(x)dx= ψ*ψ dx = (2/L) sin2(3πx/L)dx
Solve
P(x) = 5 nm-1

so set the integral equal to 5nm-1 right?

 

[gv3]

After integrating i get 5=(x/L)-(1/6π)sin(6πx/L) How do you even solve for x?

 

[kuruman]

You don't integrate.. If you integrate x drops out and you get just a number, i.e. the probability that the particle be found within the limits of integration. You need to find at what values of x the probability density[/color] has the given value. It looks like you are confusing "probability" with "probability density". They are not the same.

 

[gv3]

You don't integrate.. If you integrate x drops out and you get just a number, i.e. the probability that the particle be found within the limits of integration. You need to find at what values of x the probability density has the given value. It looks like you are confusing "probability" with "probability density". They are not the same.

Ok I got it. Thank you. The answer was .21 nm. Originally when I first solved it I got .021nm and the homework said I was off by a power of ten. Do you know why it could be off? Also could you explain to me why question 2 is 6 values of x

 

[kuruman]

Do you know why it could be off?

Not unless you reproduce and show me what you did.

Also could you explain to me why question 2 is 6 values of x

Plot P(x) as a function of x from 0 to L and you will see.

 

[gv3]

You have all you need.
P(x)dx= ψ^*ψ dx = (2/L) sin²(3πx/L)dx
Solve
P(x) = 5 nm^-1

I solved this and couldn't get 6. i tried with my calculator in radians and degrees.

 

[kuruman]

Here is the plot of P(x) vs. x. Note that the 5 nm^-1 line crosses the plot at 6 points.

 

[gv3]

Here is the plot of P(x) vs. x. Note that the 5 nm^-1 line crosses the plot at 6 points.

View attachment 196861

Ah ok. I was graphing the wrong thing. Thanks for helping me! Sorry for the confusion.