Question

Determine the probability that a particle in 1-dimension box of length a, with no forces acting on it, in state corresponding to n=10,will be found in the region between a/3 and 2a/3

Answer 1

`(1)/(3)`

Explanation:

From the Schrodinger equation, we have that the the wave function
`\psi` for a particle in 1-dimension box of length a, with no forces acting on it, is given by

`\psi = \sqrt{(2)/(a) } \sin \left( (n\pi x)/(a) \right)`

where

• 
  `a` is the length of the box
• 
  `n` is the energy level
• 
  `x` is a variable depending on the size of the box

In this case, we have a state corresponding to
`n = 10`. Hence,

`\psi = \sqrt{(2)/(a) } \sin \left( (10\pi x)/(a) \right)`

To find the probability that a particle will be found in a region between
`(a)/(3)` and
`(2a)/(3)` , we need to calculate the following integral

`p = \int\limits^{(a)/(3)}_{(2a)/(3)} \psi \psi ^* \, dx`

where
`\psi^*` is the conjugate wave equation of
`\psi` . Thus,

`\psi ^* = \sqrt{(2)/(a) } \sin \left( (n\pi x)/(a) \right)`

and we have

`p = \int\limits^{(a)/(3)}_{(2a)/(3)} \sqrt{(2)/(a) } \sin \left( (n\pi x)/(a) \right) \cdot \sqrt{(2)/(a) } \sin \left( (n\pi x)/(a) \right) \, dx \\\phantom{p} = (2)/(a) \int\limits^{(a)/(3)}_{(2a)/(3)} \sin ^2\left( (n\pi x)/(a) \right) \, dx`

To solve the integral above, use the trigonometric identity

`\sin ^2 x = (1 - \cos 2x)/(2)`

Therefore,

`p = (2)/(a) \int\limits^{(a)/(3)}_{(2a)/(3)} (1- \cos 2\left( (n\pi x)/(a) \right))/(2) \, dx\\\phantom{p} = (2)/(a) \int\limits^{(a)/(3)}_{(2a)/(3)} (dx)/(2) - (2)/(a) \int\limits^{(a)/(3)}_{(2a)/(3)} (\cos 2\left( (n\pi x)/(a) \right))/(2) \, dx\\\phantom{p} = (1)/(a) x \Big|\limits^{(a)/(3)}_{(2a)/(3)} \\\phantom{a} = (1)/(a) \left((2a)/(3) - (a)/(3) \right) \\\phantom{a} = (1)/(3)`