Question

An electron is confined to a region of space the sixe of an atom (0.1 nm). In the first excited state, what is the probability of finding the electron between x = 0 and x = 0.25 x 10^-10 m?

Answer 1

P(0<x<
`0.025* 10^(- 10)m`) = 0.25

Given:

l = d = 0.1 nm
`0.1* 10^(- 9) m`

Solution:

The wave function for the confined electron in the range 0<x<
`0.25* 10^(- 10) m` is given by:

`\Psi (x) = \sqrt{(2)/(d)}sin((2\pi x)/(d))`

Now, the probability to find the electron in the range 0<x<
`0.25* 10^(- 10) m` is given by:

P(x) =
`\int_(0)^(l)|\Psi(x)|^(2) dx`

P(x) =
`(2)/(d)\int_(0)^{0.25* 10^(- 10)}sin^(2)((2\pi x)/(d)) dx`

P(x) =
`(2)/(0.1* 10^(- 9))\int_(0)^{0.25* 10^(- 10)}sin^(2)((2\pi x)/(d)) dx`

P(x) =
`(2)/(0.1* 10^(- 9))\int_(0)^{0.25* 10^(- 10)}(1 - cos((4\pi x)/(d))/(2)) dx`

P(x) =
`\frac{2}{0.1* 10{- 9}}[(0.025* 10^(- 9)* 4\pi - (sin(4\pi* 0.025* 10^(- 9)))/(0.1* 10^(- 9)))/(8\pi)]`

On solving the above eqn, we get:

P(x) = 0.25

P(x) = 0.25