contestada

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

Respuesta :

Answer:

The probability density function is [tex]0.25\times 10^{8} m^{- 1}[/tex]

Solution:

Position of the particle in the box, S = [tex]10^{- 10} m[/tex]

Length of the box, L = [tex]20\times 10^{- 10} m[/tex]

Now, the probablity is given by:

P = [tex]\frac{10^{- 10}}{20\times 10^{- 10}} = \frac{1}{20}[/tex]

Now,

The probability density, [tex]\psi = \frac{P}{L}[/tex]

[tex]\psi = \frac{\frac{1}{20}}{20\times 10^{- 10}} = 0.25\times 10^{8} m^{- 1}[/tex]