The wavelength of the light is about 4.34 × 10⁻⁷ m
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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 = 2
final shell = n₂ = 5
Unknown:
ΔE = ?
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}{5^2} - \frac{1}{2^2})[/tex]
[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{25} - \frac{1}{4} )[/tex]
[tex]\Delta E = -2.18 \times 10^{-18} \times (-\frac{21}{100})[/tex]
[tex]\boxed{\Delta E \approx 4.578 \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]4.578 \times 10^{-19} = 6.63 \times 10^{-34} \times \frac{3 \times 10^8}{\lambda}[/tex]
[tex]\boxed{\lambda \approx 4.34 \times 10^{-7} \texttt{ m}}[/tex]
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Grade: College
Subject: Physics
Chapter: Quantum Physics
