Question

An electron that has an energy of approximately 3.0e+02 eV moves between rigid walls 0.6 nm apart.

(a) Find the quantum number n for the energy state that the electron occupies.
_____________

(b) Find the precise energy of the electron.
_____________eV

Answer 1

(a)

The allowed energy levels for a particle in a box are

n = 1,2,3...

Here, Planck’ constant is h, mass of the electron is m and length of the rigid wall is L.

Substitute 9.1 x 10^-31 for m, 0.6 nm for L and 6.63 x 10^-34 Js for h in the above equation,

For n = 16,

E₁₆ = (16)² (1.0471 eV)

        = 268.05 eV

For n = 17,

E₁₇ = (17)² (1.0471 eV)

        = 302.61 eV

For n = 18,

E₁₈ = (18)² (1.0471 eV)

        = 339.26 eV

It is given that the energy of the electron is 3.0e+02 or 300 eV.

From the above, it is found that the energy of the electron is quite near to energy corresponding quantum number n = 17.

Therefore, the quantum number n for the energy state that the electron occupies is n = 17.

(b)

For n =17, the energy state of the electron is

E₁₇ = 302.61 eV

Rounding off to three significant figures, the precise energy of the electron is 303 eV.