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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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a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

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where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

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The displacement is

d = 4.0 m

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b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

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Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Ksenya-84 [330]

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