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

This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3

Answer 2

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 : 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