Respuesta :
Answer:
[tex]n_f=2[/tex]
Explanation:
It is given that,
Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :
[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]
[tex]\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})[/tex]
R = Rydberg constant, [tex]R=1.097\times 10^7\ m^{-1}[/tex]
[tex]\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})[/tex]
Solving above equation we get the value of final n is,
[tex]n_f=2.04[/tex]
or
[tex]n_f=2[/tex]
So, it will relax in the n = 2. Hence, this is the required solution.
The electron relaxes to energy level n = 2
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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 )
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The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem !
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Given:
initial shell = n₁ = 7
wavelength = λ = 397 nm = 3.97 × 10⁻⁷ m
Unknown:
final shell = n₂ = ?
Solution:
We will use this following formula to solve this problem:
[tex]\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]
[tex]h \frac{c}{\lambda} = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]
[tex]6.63 \times 10^{-34} \frac{3 \times 10^8}{3.97 \times 10^{-7}} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{7^2})[/tex]
[tex]5.01008 \times 10^{-19} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{7^2})[/tex]
[tex]( \frac{1}{(n_2)^2} - \frac{1}{7^2}) = \frac{45}{196}[/tex]
[tex]\frac{1}{(n_2)^2} = \frac{1}{49} + \frac{45}{196}[/tex]
[tex]\frac{1}{(n_2)^2} = \frac{1}{4}[/tex]
[tex]n_2)^2 = 4[/tex]
[tex]n_2 = \sqrt{4}[/tex]
[tex]\boxed{n_2 = 2}[/tex]
[tex]\texttt{ }[/tex]
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
