Question

If the particle is in the ground state, what is the probability that it is in a window Δx=0.0002L wide with its midpoint at x=0.700L? You should be able to answer this part without evaluating any integrals! Since Δx is so small, you can assume that ψ(x) remains constant over that interval, so the integral is approximately

Answer 1

To solve this problem it is necessary to apply the concepts related to the wave function of a particle and the probability of finding the particle in the ground state.

The wave function is given as

$\phi = \sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})$

Therefore the probability of finding the particle must be

$P = |\phi(x)^2| \Delta X$

$P = |\sqrt{\frac{2}{L}} sin(\frac{\pi x}{L})|^2 \Delta x$

$P = (\frac{2}{L})(sin(\frac{\pi x}{L}))^2\Delta x$

$P = (\frac{2}{L})(sin(\frac{\pi (0.7L)}{L}))^2 (0.0002L)$

$P = 2*(sin(0.7\pi))^2(0.0002)$

$P = 2.618*10^{-4}$

Therefore the probability is $2.618*10^{-4}$