Respuesta :
The order of frequency is d < a < c < b
[tex]E_{n} = -13.6 * z^{2} / n^{2}[/tex] eV
where z = atomic mass number
n = energy level
For hydrogen z = 1
The expression for the energy of the electron is:
E = -13.6 * Z^2 / n^2
Therefore, Energy for n = 1
[tex]E_{n} = -13.6 * 1^{2} / 1^{2}[/tex]
= -13.6 eV
for n = 2
[tex]E_{n} = -13.6 * 1^{2} / 2^{2}[/tex]
= -3.40 eV
for n = 3
[tex]E_{n} = -13.6 * 1^{2} / 3^{2}[/tex]
= -1.51 eV
for n = 4
[tex]E_{n} = -13.6 * 1^{2} / 4^{2}[/tex]
= -0.85 eV
for n = 5
[tex]E_{n} = -13.6 * 1^{2} / 5^{2}[/tex]
= -0.544 eV
n = 2 to n = 4 (absorption) since the transition is from lower to higher level energy is required.
ΔE = [tex]E_{4}[/tex] - [tex]E_{2}[/tex] = -0.85 - (-3.40) = 2.55 eV
n = 2 to n = 1 (emission) since transition is from higher to lower level energy is released.
ΔE = [tex]E_{1}[/tex] - [tex]E_{2}[/tex] = -13.6 - (-3.40) = -10.2eV
The negative sign indicates that emission will take place.
n = 2 to n = 5 (absorption)
ΔE = [tex]E_{5}[/tex] - [tex]E_{2}[/tex] = -0.544 - (-3.40) = 2.856 eV
n = 4 to n = 3 (emission)
ΔE = [tex]E_{3}[/tex] - [tex]E_{4}[/tex] = -1.51 - (-0.85) = -0.66 eV
We know that
E = h * υ
where υ = frequency
Therefore, Energy is proportional to frequency.
So increasing the order of energy is
[tex]E_{4}[/tex] < [tex]E_{1}[/tex] < [tex]E_{3}[/tex] < [tex]E_{2}[/tex]
order of frequency is
d < a < c < b
For more information on the frequency of photons click on the link below:
https://brainly.com/question/17058029
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