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trasher [3.6K]
3 years ago
10

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

of n for the level to which the electron relaxed?

Physics
2 answers:
Dimas [21]3 years ago
6 0

Answer:

n_f=2

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

R = Rydberg constant, R=1.097\times 10^7\ m^{-1}

\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})

Solving above equation we get the value of final n is,

n_f=2.04

or

n_f=2

So, it will relax in the n = 2. Hence, this is the required solution.        

Anika [276]3 years ago
4 0

The electron relaxes to energy level n = 2

\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>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{ }

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.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

initial shell = n₁ = 7

wavelength = λ = 397 nm = 3.97 × 10⁻⁷ m

<u>Unknown:</u>

final shell = n₂ = ?

<u>Solution:</u>

<em>We will use this following formula to solve this problem:</em>

\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})

h \frac{c}{\lambda} = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})

6.63 \times 10^{-34} \frac{3 \times 10^8}{3.97 \times 10^{-7}} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{7^2})

5.01008 \times 10^{-19} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{7^2})

( \frac{1}{(n_2)^2} - \frac{1}{7^2}) = \frac{45}{196}

\frac{1}{(n_2)^2} = \frac{1}{49} + \frac{45}{196}

\frac{1}{(n_2)^2} = \frac{1}{4}

n_2)^2 = 4

n_2 = \sqrt{4}

\boxed{n_2 = 2}

\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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