Answer:
The energies in descending order 35.3 eV, 28.2 eV, 18.8 eV, 16.5 eV, 11.8 eV, 7.05 eV.
Explanation:
Given that,
Width = 0.4 nm
Number of state n= 4
We need to calculate the energy in first state
When n = 1,
Using formula of energy
[tex]E_{1}=\dfrac{h^2}{8mL^2}[/tex]
Put the value into the formula
[tex]E_{1}=\dfrac{(6.63\times10^{-34})^2}{8\times9.1\times10^{-31}\times(0.4\times10^{-9})^2}[/tex]
[tex]E_{1}=3.77\times10^{-19}\ J[/tex]
[tex]E_{1}=\dfrac{3.77\times10^{-19}}{1.6\times10^{-19}}[/tex]
[tex]E_{1}=2.35\ eV[/tex]
Energy in nth level is
[tex]E_{n}=n^2\times E_{1}[/tex]
Photon energy in different states is
[tex]E_{p}=E_{n}-E_{n-1}[/tex]
Put the value into the formula
[tex]E_{4-3}=(4^2-3^2)\times2.35=16.5\ eV[/tex]
[tex]E_{4-2}=(4^2-2^2)\times2.35=28.2\ eV[/tex]
[tex]E_{4-1}=(4^2-1^2)\times2.35=35.3\ eV[/tex]
[tex]E_{3-2}=(3^2-2^2)\times2.35=11.8\ eV[/tex]
[tex]E_{3-1}=(3^2-1^2)\times2.35=18.8\ eV[/tex]
[tex]E_{2-1}=(2^2-1^2)\times2.35=7.05\ eV[/tex]
Hence, The energies in descending order 35.3 eV, 28.2 eV, 18.8 eV, 16.5 eV, 11.8 eV, 7.05 eV.
