Probability of electron in a box problem

[TLeo198]

Homework Statement

Calculate the probability that an electron will be found (a) between x = 0.1 and 0.2 nm, (b) between 4.9 and 5.2 nm in a box of length L = 10 nm when its wavefunction is psi = (2/L)^1/2 sin(2*pi*x/L).

Homework Equations

As far as I know, only the normalization equation, which is A^2 * Integral(psi^2) = 1

The Attempt at a Solution

My problem is actually how to start it. Should I assume that the wavefunction is already normalized, or should I normalize the wavefunction by equating the integral of the squared wavefunction to 1, then solve for A? Sorry if wording of the question is bad. Any help is appreciated.

 

[zhermes]

It looks like the wavefunction is already normalized. Its easy to check.

The fundamental principle of the probabilistic interpretation of quantum mechanics is that the probability of finding a particle between points a and b, is equal to the integral of the wave-function squared over that interval:

$$P(a<x<b) = \int_a^b |\Psi(x) |^2 dx$$

 

[TLeo198]

I was able to solve it, thanks!

 

[mspns]

Given two normalized but non-orthogonal eigen functions, psi=(1/sqrt(pi))exp(-r) and phi=(1/sqrt(3pi))rexp(-r). Construct a new function PSI which is orthogonal to the first function and is normalized.

Can anyone help me please?


Many many thanks for your kind help.

MSPNS

 

[paranormal]

I would approach this problem by first understanding the concept of probability in quantum mechanics. In quantum mechanics, the probability of finding a particle at a certain position is given by the squared magnitude of its wavefunction at that position. Therefore, to calculate the probability of finding an electron in a certain range of positions, we need to integrate the squared wavefunction over that range.

For part (a), we need to integrate the squared wavefunction from x = 0.1 nm to x = 0.2 nm. This can be done by breaking up the integral into two parts, from x = 0.1 nm to x = 0.2 nm and from x = 0 to x = 0.1 nm, and then subtracting the second part from the first. This is because the wavefunction is zero for x < 0 nm.

The integral for the first part would be A^2 * Integral(sin^2(2*pi*x/L)) from x = 0.1 nm to x = 0.2 nm. This can be simplified using trigonometric identities to give A^2 * (0.1 - 0.2 + (sin(4*pi/10) - sin(8*pi/10)) / (4*pi)). Similarly, the integral for the second part would be A^2 * Integral(sin^2(2*pi*x/L)) from x = 0 to x = 0.1 nm, which simplifies to A^2 * (sin(2*pi/10) - sin(8*pi/10)) / (4*pi).

Subtracting the second part from the first, we get A^2 * (0.1 - (sin(2*pi/10) - sin(8*pi/10)) / (4*pi)). Since we know that the integral of the squared wavefunction over all space should be equal to 1, we can set this expression equal to 1 and solve for A. This gives us A = (2*pi)^1/2.

We can then use this value of A to calculate the probability of finding the electron between x = 0.1 and 0.2 nm by plugging it into the integral expression and evaluating it. This gives us a probability of approximately 0.077.

For part (b), the process is similar. We need to integrate the squared wavefunction from x = 4.9 nm to