If a particle in a box has a ground-state energy of $10.0 \mathrm{eV}$, what is the energy of the fourth excited state?
A) $250 \mathrm{eV}$
B) $160 \mathrm{eV}$
C) $50.0 \mathrm{eV}$
D) $20.0 \mathrm{eV}$
E) $40.0 \mathrm{eV}$

Question

Quizplus · Accepted Answer

The energy of the nth excited state is given by:$$E_n = n^2 * E_1$$where $E_1$ is the ground-state energy. Therefore, the energy of the fourth excited state is:$$E_4 = 4^2 * 10.0 \mathrm{eV} = 160 \mathrm{eV}$$

If a Particle in a Box Has a Ground-State Energy 10.0eV10.0 \mathrm{eV}10.0eV

If a Particle in a Box Has a Ground-State Energy $10.0 \mathrm{eV}$