The electron moves to energy level n = 3
[tex]\texttt{ }[/tex]
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 )
[tex]\texttt{ }[/tex]
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 !
[tex]\texttt{ }[/tex]
Given:
initial shell = n₁ = 5
wavelength = λ = 1282.17 nm = 1.28217 × 10⁻⁶ 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} \times \frac{3 \times 10^8}{1.28217 \times 10^{-6}} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{5^2})[/tex]
[tex]1.55128 \times 10^{-19} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{5^2})[/tex]
[tex]( \frac{1}{(n_2)^2} - \frac{1}{5^2}) = \frac{16}{225}[/tex]
[tex]\frac{1}{(n_2)^2} = \frac{1}{25} + \frac{16}{225}[/tex]
[tex]\frac{1}{(n_2)^2} = \frac{1}{9}[/tex]
[tex](n_2)^2 = 9[/tex]
[tex]n_2 = \sqrt{9}[/tex]
[tex]\boxed{n_2 = 3}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: College
Subject: Physics
Chapter: Quantum Physics
The electron will move to energy level n=3.
Using the Rydberg formula;
1/λ = R(1/nf^2 - 1/ni^2)
Where;
λ = 1282.17 nm or 1282.17 × 10^-9 m
nf = ?
ni = 5
R = 1.097 × 10^7 m-1
Substituting the values;
1/1282.17 × 10^-9 m = 1.097 × 10^7 m-1(1/nf^2 - 1/5^2)
0.0713 = 1/nf^2 - 1/5^2
1/nf^2 = 0.0713 + 0.04
nf = 3
The electron moves to energy level 3.
Learn more: https://brainly.com/question/11969651
