Question

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?

Answer 1

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.        

Answer 2

The electron relaxes to energy level n = 2

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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 :

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

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.

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

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

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 = 7

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

Unknown:

final shell = n₂ = ?

Solution:

We will use this following formula to solve this problem:

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

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

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

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

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

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

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

$n_2)^2 = 4$

$n_2 = \sqrt{4}$

$\boxed{n_2 = 2}$

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Learn more

• Photoelectric Effect : brainly.com/question/1408276
• Statements about the Photoelectric Effect : brainly.com/question/9260704
• Rutherford model and Photoelecric Effect : brainly.com/question/1458544

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Answer details

Grade: College

Subject: Physics

Chapter: Quantum Physics