Respuesta :
Answer:
P(0<x< [tex]0.025\times 10^{- 10}m[/tex]) = 0.25
Given:
l = d = 0.1 nm [tex]0.1\times 10^{- 9} m[/tex]
Solution:
The wave function for the confined electron in the range 0<x<[tex]0.25\times 10^{- 10} m[/tex] is given by:
[tex]\Psi (x) = \sqrt{\frac{2}{d}}sin(\frac{2\pi x}{d})[/tex]
Now, the probability to find the electron in the range 0<x<[tex]0.25\times 10^{- 10} m[/tex] is given by:
P(x) = [tex]\int_{0}^{l}|\Psi(x)|^{2} dx[/tex]
P(x) = [tex]\frac{2}{d}\int_{0}^{0.25\times 10^{- 10}}sin^{2}(\frac{2\pi x}{d}) dx[/tex]
P(x) = [tex]\frac{2}{0.1\times 10^{- 9}}\int_{0}^{0.25\times 10^{- 10}}sin^{2}(\frac{2\pi x}{d}) dx[/tex]
P(x) = [tex]\frac{2}{0.1\times 10^{- 9}}\int_{0}^{0.25\times 10^{- 10}}\frac{1 - cos(\frac{4\pi x}{d}}{2}) dx[/tex]
P(x) = [tex]\frac{2}{0.1\times 10{- 9}}[\frac{0.025\times 10^{- 9}\times 4\pi - \frac{sin(4\pi\times 0.025\times 10^{- 9})}{0.1\times 10^{- 9}}}{8\pi}] [/tex]
On solving the above eqn, we get:
P(x) = 0.25
P(x) = 0.25
