De Broglie wavelength of an electron in the ground state of the hydrogen atom is about 3.33 × 10⁻¹⁰ m

<h3>Further explanation</h3>
The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :

<em>where:</em>
<em>E = Energi of A Photon ( Joule )</em>
<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>
<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

Let's recall De Broglie's Wavelength Formula as follows:

<em>where:</em>
<em>λ = wavelength ( m )</em>
<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>
<em>m = mass of object ( kg )</em>
<em>v = velocity of object ( m/s )</em>
Let us now tackle the problem !

<u>Given:</u>
energy of the ground state of the hydrogem atom = E = 13.6 eV = 2.176 × 10⁻¹⁸ J
<u>Asked:</u>
wavelength of electron = λ = ?
<u>Solution:</u>
<em>Firstly , we will calculate the speed of the electron :</em>



<em>→ Equation A</em>

<em>Next, we will use the formula of The Broglie's Wavelength:</em>

←<em> Equation A</em>



<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: College
Subject: Physics
Chapter: Quantum Physics