Hello!

Yes, this probability is an integral from `12 - 0.2 = 11.8 nm` to 1`2 + 0.2 = 12.2 nm.` There are two problems: what is the function to integrate and what is the value of the integral.

It is known that the function to integrate is the probability density, and it is the square of the wave function. Also it is known that the wave function for a particle in an infinite one-dimensional well with the walls `x=0` and `x=L` is

`Psi(x) = sqrt(2/L) sin((n pi x)/L),`

where `n` is the state. So we need to integrate

`p_d(x) = 2/Lsin^2((n pi x)/L) = 1/L (1-cos((2n pi x)/L)).`

It is simple, and the probability is

`int_(11.8)^(12.2) 1/L (1-cos((2n pi x)/L)) dx = (x/L - 1/(2n pi) sin((2n pi x)/L))|_(x=11.8)^(12.2)`

Recall that `n=2` and `L=15nm` and obtain

`p = 0.4/15-1/(4pi)(sin(4pi*12.2/15)-sin(4pi*11.8/15)) approx 0.048.`

(that said, your 0.027 is the first term, 0.4/15)

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now