Question

An electron is trapped in an infinite square-well potential of width 0.6 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? (List in descending order of energy. Enter 0 in any remaining unused boxes.)

Answer 1

E₁ = 1.042 eV

E₄₋₃= 7.29 eV

E₄₋₂= 12.50 eV

E₄₋₁= 15.63 eV

E₃₋₂= 5.21eV

E₃₋₁= 8.34eV

E₂₋₁= 3.13eV

Step-by-step explanation:

The energy in an infinite square-well potential is giving by:

`E = \frac {h^(2) n^(2)}{8mL^(2)}`

where, h: Planck constant = 6.62x10⁻³⁴J.s, n: is the energy state, m: mass of the electron and L: widht of the square-well potential

The energy of the electron in the ground state, n = 1, is:

`E_(1) = \frac {(6.62 \cdot 10^(-34))^(2) (1)^(2)}{(8) (9.11 \cdot 10^(-31)) (0.6 \cdot 10^(-9) m)^(2)}`

`E_(1) = 1.67 \cdot 10^(-19) J = 1.042 eV`

The photon energies that are emitted as the electron jumps to the ground state is the difference between the states:

`E_(\Delta n) = \Delta n^(2) E_(1)`

`E_((4 - 3)) = (4^(2) - 3^(2)) 1.042 eV = 7.29eV`

`E_((4 - 2)) = (4^(2) - 2^(2)) 1.042 eV = 12.50eV`

`E_((4 - 1)) = (4^(2) - 1^(2)) 1.042 eV = 15.63eV`

`E_((3 - 2)) = (3^(2) - 2^(2)) 1.042 eV = 5.21eV`

`E_((3 - 1)) = (3^(2) - 1^(2)) 1.042 eV = 8.34eV`

`E_((2 - 1)) = (2^(2) - 1^(2)) 1.042 eV = 3.13eV`

Have a nice day!