Question

An electron is trapped in an infinite square-well potential of width 0.70 nm. If the electron is initially in the n=4 state, what are the various photon energies that can be emitted as the electron jumps to the ground state, considering that the electron doesn not have to go directly to the ground state but can make an intermediate step (e.g. from n=4 to n=2 to n=1)

Answer 1

Photons with different energy are

12.9 10⁻²⁸ J , 6.04 10⁻²⁸ J , 6.9 10⁻²⁸ J, 10.35 10⁻²⁸J, 2,588 10⁻²⁸J

Step-by-step explanation:

The solution of the Schrödinger equation for an infinite square potential is relatively easy and results in the energy having the form

Eₙ = (h²/8mL²) n²

Where m is the mass of the electron, L the length of the box and n is a positive integer

Let's apply this equation the case that they give us

L = 0.70 nm = 0.70 10-9 m

E₀ = h² / 8mL²

E₀ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ 0.70 10⁻⁹)

E₀ = 0.8626 10⁻²⁸ J

Eₙ = E₀ n²

For the initial state n = 4

E₄ = Eo 4²

E₄ = 16 E₀

Let's write the possible transitions

Case first jump second jump

State: initial final initial final

1 4 1

2 4 3 3 1

3 4 2 2 1

Energy of emitted photons

Case 1

ΔE = E₄ -E₁

DE = 16 E₀ - Eo = 15 E₀

ΔE = 15 0.8626 10⁻²⁸ J

ΔE = 12.9 10⁻²⁸ J

Case 2

First photon

ΔE = E₄ -E₃

ΔE = 16E₀ - 9E₀ = 7 E₀

ΔE = 7 0.8626 10⁻²⁸ J

ΔE = 6.04 10⁻²⁸ J

Photon 2

ΔE = E₃-E₁

ΔE = 9E₀ - E₀ =8E₀

ΔE = 8 0.8626 10⁻²⁸ J

ΔE = 6.9 10⁻²⁸ J

Case 3

Photon 1

ΔE = E₄-E₂

ΔE = 16E₀-4E₀ = 12E₀

ΔE = 12 0.8626 10⁻²⁸ J

ΔE = 10.35 10⁻²⁸ J

Photon 2

ΔE = E₂-E₁

ΔE = 4E₀ -E₀ = 3E₀

ΔE = 3 0.8626 10⁻²⁸ J

ΔE = 2,588 10⁻²⁸ J

Photons with different energy are

- 12.9 10⁻²⁸ J

- 6.04 10⁻²⁸ J

- 6.9 10⁻²⁸ J

- 10.35 10⁻²⁸J

- 2,588 10⁻²⁸J

In total 5 photons of different energy are emitted