Complete Question
(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .
(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.
Answer:
A
[tex]\Delta E = 337 \ eV[/tex]
B
[tex]\lambda = 3.439 *10^{-9} \ m[/tex]
Explanation:
Considering question a
From the question we are told that
The width of the box is [tex]w = 0.125 \ nm = 0.125 *10^{-9} \ m[/tex]
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ][/tex]
=> [tex]\Delta E = 539 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J[/tex]
=> [tex]\Delta E = 337 \ eV[/tex]
Considering question b
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 4 \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ][/tex]
=> [tex]\Delta E = 578 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}[/tex]
=> [tex]\Delta E = 361.45 \ eV[/tex]
Gnerally the wavelength is mathematically represented as
[tex]\lambda = \frac{hc}{\Delta E }[/tex]
=> [tex]\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }[/tex]
=> [tex]\lambda = 3.439 *10^{-9} \ m[/tex]
