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

Answer:

The probability density function is $0.25\times 10^{8} m^{- 1}$

Solution:

Position of the particle in the box, S = $10^{- 10} m$

Length of the box, L = $20\times 10^{- 10} m$

Now, the probablity is given by:

P = $\frac{10^{- 10}}{20\times 10^{- 10}} = \frac{1}{20}$

Now,

The probability density, $\psi = \frac{P}{L}$

$\psi = \frac{\frac{1}{20}}{20\times 10^{- 10}} = 0.25\times 10^{8} m^{- 1}$