Question

calculate lowest energy of the system containing two electrons confined to a box of length one Armstrong.​

Answer 1

Answer:

· The potential energy is 0 inside the box (V=0 for 0<x<L) and goes to infinity at the walls of the box (V=∞ for x<0 or x>L). We assume the walls have infinite potential energy to ensure that the particle has zero probability of being at the walls or outside the box.

Answer 2

The lowest energy of the system containing two electrons confined to a box of length one Armstrong is 24.12aJ.

Explanation:

Given,

2 electrons in a box of length 1 Armstrong.

To find,

Lowest energy of the system.

Concept used,

By using concepts of quantum mechanics,

$E = \frac{h^{2} }{8ma^{2} }$.

The state of lowest energy states is also called as ground state.

$E = \frac{h^{2} }{8ma^{2} }$

If we substitute planks constant in the formula and other values we would get and answer.

$E = \frac{h^{2} }{8ma^{2} } = 24.12aJ.$

Therefore the lowest energy of the system containing two electrons confined to a box of length one Armstrong is 24.12aJ.