Respuesta :
The Rydberg formula was formulated by Johannes Rydberg for the calculation of the wavelength of an atomic element when an excited electron moves from the quantum shells. The formula is
1/wavelength=R(1/n1^2 - 1/n2^2), where n2 and n1 are the principal quantum numbers and R is the Rydberg constant which is equal to 1.097 x 107 m−1.
As mentioned, the electron moves from an excited state n2, to the ground-level state n1. Therefore, n2 is always greater than n1. Substituting the values to the equation:
1/wavelength=1.097 x 10^7 (1/2^2 - 1/4^2)
wavelength = 4.8617 meters or 48.6 micrometers
1/wavelength=R(1/n1^2 - 1/n2^2), where n2 and n1 are the principal quantum numbers and R is the Rydberg constant which is equal to 1.097 x 107 m−1.
As mentioned, the electron moves from an excited state n2, to the ground-level state n1. Therefore, n2 is always greater than n1. Substituting the values to the equation:
1/wavelength=1.097 x 10^7 (1/2^2 - 1/4^2)
wavelength = 4.8617 meters or 48.6 micrometers
The wavelength of the spectral lline produced is about 4.87 × 10⁻⁷ m
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Further explanation
The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
Let us now tackle the problem !
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Given:
initial shell = n₁ = 4
final shell = n₂ = 2
Asked:
λ = ?
Solution:
Firstly, we will use this following formula to calculate the change in energy of the electron:
[tex]\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]
[tex]\Delta E = 2.18 \times 10^{-18} \times ( \frac{1}{2^2} - \frac{1}{4^2})[/tex]
[tex]\Delta E = 2.18 \times 10^{-18} \times ( \frac{1}{4} - \frac{1}{16} )[/tex]
[tex]\Delta E = 2.18 \times 10^{-18} \times \frac{3}{16}[/tex]
[tex]\boxed{\Delta E \approx 4.0875 \times 10^{-19} \texttt{ J}}[/tex]
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Next, we will calculate the wavelength of the light:
[tex]\Delta E = h \frac{c}{\lambda}[/tex]
[tex]4.0875 \times 10^{-19} = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda}[/tex]
[tex]\boxed{\lambda \approx 4.87 \times 10^{-7} \texttt{ m}}[/tex]
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Learn more
- Photoelectric Effect : https://brainly.com/question/1408276
- Statements about the Photoelectric Effect : https://brainly.com/question/9260704
- Rutherford model and Photoelecric Effect : https://brainly.com/question/1458544
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Answer details
Grade: College
Subject: Physics
Chapter: Quantum Physics
