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

Answer:

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

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!