De Broglie wavelength of an electron in the ground state of the hydrogen atom is about 3.33 × 10⁻¹⁰ 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]
where:
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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Let's recall De Broglie's Wavelength Formula as follows:
[tex]\boxed{\lambda = \frac{h}{mv}}[/tex]
where:
λ = wavelength ( m )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
m = mass of object ( kg )
v = velocity of object ( m/s )
Let us now tackle the problem !
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Given:
energy of the ground state of the hydrogem atom = E = 13.6 eV = 2.176 × 10⁻¹⁸ J
Asked:
wavelength of electron = λ = ?
Solution:
Firstly , we will calculate the speed of the electron :
[tex]E = E_k[/tex]
[tex]E = \frac{1}{2}mv^2[/tex]
[tex]v^2 = 2E \div m[/tex]
[tex]\boxed{v = \sqrt{ 2E \div m } }[/tex] → Equation A
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Next, we will use the formula of The Broglie's Wavelength:
[tex]\lambda = \frac{h}{mv}[/tex]
[tex]\lambda = \frac{h}{m\sqrt{ 2E \div m }}[/tex] ← Equation A
[tex]\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \sqrt{ 2 \times 2.176 \times 10^{-18} \div 9.11 \times 10^{-31} }}[/tex]
[tex]\boxed{\lambda = 3.33 \times 10^{-10} \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
