Question

Find the probability density of a particle moving in an interval of 101o the box. The length of the one-dimensional box is 20x10-10 m (a) 0.2 (b) 0.3 (c) 0.4 (d) None of the above

Answer 1

The probability density function is
`0.25* 10^(8) m^(- 1)`

Solution:

Position of the particle in the box, S =
`10^(- 10) m`

Length of the box, L =
`20* 10^(- 10) m`

Now, the probablity is given by:

P =
`(10^(- 10))/(20* 10^(- 10)) = (1)/(20)`

Now,

The probability density,
`\psi = (P)/(L)`

`\psi = ((1)/(20))/(20* 10^(- 10)) = 0.25* 10^(8) m^(- 1)`