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OLEGan [10]
3 years ago
8

Find the de broglie wavelength of an electron in the ground state of the hydrogen atom.

Physics
2 answers:
HACTEHA [7]3 years ago
5 0

de Broglie wavelength (λ) is given by the equation

λ = h/p

where h=Planck’s constant whose value is 6.62 x 10^(−34) joule-seconds and

p = momentum of the particle(here electron)

In terms of kinetic energy(E) momentum(p) can be written as,

p=(2mE)^1/2

where m=mass of the particle.

Hence λ becomes

1 λ = h(2mE)^-1/2

Given here, E = 13.6 eV = 13.6×1.6×10^-19 joule

m(mass of electron)= 9.1×10^-31 kg

Putting these values in equation (1) we get ,

λ =0.332×10^(-9) meter

=3.32×10^(-10) meter

=3.32 Å

yanalaym [24]3 years ago
4 0

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

\texttt{ }

<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 :

\large {\boxed {E = h \times f}}

<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>

\texttt{ }

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

\boxed{\lambda = \frac{h}{mv}}

<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 !

\texttt{ }

<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>

E = E_k

E = \frac{1}{2}mv^2

v^2 = 2E \div m

\boxed{v = \sqrt{ 2E \div m } } <em>→ Equation A</em>

\texttt{ }

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

\lambda = \frac{h}{mv}

\lambda = \frac{h}{m\sqrt{ 2E \div m }} ←<em> Equation A</em>

\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} }}

\boxed{\lambda = 3.33 \times 10^{-10} \texttt{ m}}

\texttt{ }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544

\texttt{ }

<h3>Answer details</h3>

Grade: College

Subject: Physics

Chapter: Quantum Physics

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