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Determine the energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom. determine the energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom. -1.82 × 10-19 j +2.69 × 10-19 j +5.51 × 10-19 j +3.03 × 10-19 j -3.03 × 10-19 j

Respuesta :

skyluke89
The expression for the change in energy for a transition from a state [tex]n_i[/tex] to a state [tex]n_f[/tex] in a hydrogen atom is
[tex]\Delta E=-13.6 Z^2 ( \frac{1}{n_f^2}- \frac{1}{n_i^2} )[eV][/tex]
where Z is the atomic number, which is 1 for hydrogen. Substituting [tex]n_i=3[/tex] and [tex]n_f=2[/tex] we find
[tex]\Delta E = -1.89 eV[/tex]
And to convert it into Joule, we should multiply this number by the elementary charge, obtaining:
[tex]\Delta E = (-1.89 eV)(1.6 \cdot 10^{-19}c)=-3.03 \cdot 10^{-19}C[/tex]
johanrusli

The energy change associated with the transition from n = 3 to n = 2 in the hydrogen atom is about -3.03 × 10⁻¹⁹ J

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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]

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 = 3

final shell = n₂ = 2

Unknown:

ΔE = ?

Solution:

We will use this following formula to solve this problem:

[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}{2^2} - \frac{1}{3^2})[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times ( \frac{1}{4} - \frac{1}{9} )[/tex]

[tex]\Delta E = -2.18 \times 10^{-18} \times \frac{5}{36}[/tex]

[tex]\Delta E \approx -3.03 \times 10^{-19} \texttt{ J}[/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

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