Question

An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move?

Answer 1

The electron moves to energy level n = 3

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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 = 5

wavelength = λ = 1282.17 nm = 1.28217 × 10⁻⁶ m

Unknown:

final shell = n₂ = ?

Solution:

We will use this following formula to solve this problem:

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

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

$6.63 \times 10^{-34} \times \frac{3 \times 10^8}{1.28217 \times 10^{-6}} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{5^2})$

$1.55128 \times 10^{-19} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{5^2})$

$( \frac{1}{(n_2)^2} - \frac{1}{5^2}) = \frac{16}{225}$

$\frac{1}{(n_2)^2} = \frac{1}{25} + \frac{16}{225}$

$\frac{1}{(n_2)^2} = \frac{1}{9}$

$(n_2)^2 = 9$

$n_2 = \sqrt{9}$

$\boxed{n_2 = 3}$

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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

Answer 2

The electron will  move to energy level n=3.

Using the Rydberg formula;

1/λ = R(1/nf^2 - 1/ni^2)

Where;

λ = 1282.17 nm or 1282.17 × 10^-9 m

nf = ?

ni = 5

R = 1.097 × 10^7 m-1

Substituting the values;

1/1282.17 × 10^-9 m = 1.097 × 10^7 m-1(1/nf^2 - 1/5^2)

0.0713 = 1/nf^2 - 1/5^2

1/nf^2 = 0.0713 +  0.04

nf = 3

The electron moves to energy level 3.

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