Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequency of 8.88×1014 hz . what is the change in energy, δe, that the electron undergoes to decay to its ground state? express your answer in electron volts to three significant figures.

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22-04-2018
Physics

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Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequency of 8.88×1014 hz . what is the change in energy, δe, that the electron undergoes to decay to its ground state? express your answer in electron volts to three significant figures.

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skyluke89 skyluke89 · Answer 1 · 2018-05-01T09:36:02+02:00

Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
[tex]E=hf[/tex]
where
h is the Planck constant
f is the photon frequency
Substituting [tex]f=8.88 \cdot 10^{14}Hz[/tex], we find
[tex]E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J[/tex]

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
[tex]\Delta E= -5.86 \cdot 10^{-19} J[/tex]

johanrusli johanrusli · Answer 2 · 2019-12-16T03:20:56+01:00

The change in energy is about 3.68 eV

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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 )

[tex]\texttt{ }[/tex]

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 !

[tex]\texttt{ }[/tex]

Given:

f = 8.88 × 10¹⁴ Hz

h = 6.63 × 10⁻³⁴ Js

Unknown:

ΔE = ?

Solution:

[tex]\Delta E = h \times f[/tex]

[tex]\Delta E = (6.63 \times 10^{-34}) \times (8.88 \times 10^{14})[/tex]

[tex]\Delta E = 5.88744 \times 10^{-19} \texttt{ Joule}[/tex]

[tex]\Delta E = 5.88744 \times 10^{-19} \div (1.6 \times 10^{-19})\texttt{ eV}[/tex]

[tex]\Delta E \approx 3.68 \texttt{ eV}[/tex]

[tex]\texttt{ }[/tex]

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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Keywords: Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light