An electron is confined in a one-dimensional infinite potential well of 1.0 nm.Calculate the energy difference between the 2nd excited state and the ground state A)0.8 * 10^-20 J B)2.8 * 10^-21 J C)4.8 * 10^-19 J D)3.8 * 10^-17 J

Question

Quizplus · Accepted Answer

The energy levels of a particle in a one-dimensional infinite potential well are given by:
E_n = (n^2 * h^2) / (8 * m * L^2), where n is the quantum number, h is Planck's constant, m is the mass of the particle and L is the width of the well.
For the ground state (n=1), we have:
E_1 = (1^2 * h^2) / (8 * m * L^2) = 6.107 eV
For the 2nd excited state (n=3), we have:
E_3 = (3^2 * h^2) / (8 * m * L^2) = 54.963 eV
The energy difference between the two states is therefore:
Delta E = E_3 - E_1 = 48.856 eV = 4.8 x 10^(-19) J, which corresponds to option C.

An Electron Is Confined in a One-Dimensional Infinite Potential Well