Question

an electron in the 3rd shell of an Aluminium atom moves to the first shell in a bombardment process. Calculate the frequency of the electron the 3rd orbit and energy used in transition to the first shell.[h=6.626*10^-14J/Hz [R=3.29*10^15 Hz]

Answer 1

Following the quantic theory, the energy of a photon equals the radiation frequency multiplied by the universal constant. ν = 2.923x10¹⁵ Hz. E = 3.09x10¹⁵Hz.

What is quantum mechanic?

It is the branch of physics that studies objects and forces at a very low scale, at atoms, subatoms, and particles levels.

Quantum mechanics states that the elemental particles that constitute matter -electrons, neutrons, protons- have the properties of a wave and a particle.

It emerges from the quantic theory exposed by Max Planck (1922), in which he affirmed that light propagates in energy packages or photons.

He discovered the Universal Planck constant, h, used to calculate the energy of a photon.

He stated that the energy of a photon (E) equals the radiation frequency (ν) multiplied by the universal constant (h).

E = νh

In the exposed example, we need to calculate the energy required to change from the 3rd shell to the first shell.

To do it, we should know that the energy in a level (Eₙ) equals the energy associated to an electron in the most inferior energy level (E₁) divided by the square of the shell number (n²).

Eₙ = E₁ / n²

E₁ is a constant. We can express it in Joules or electroVolts

• E₁ = -2.18x10⁻¹⁸ J
• E₁ = -13.6 eV

So, let us calculate the energy at level 1 and 3

Eₙ = E₁ / n²

• E₁ =  -2.18x10⁻¹⁸ J / 1² =  -2.18x10⁻¹⁸ J

        E₁ =  -13.6 eV / 1² =  -13.6 eV

• E₃ =  -2.18x10⁻¹⁸ J / 3² =  -2.18x10⁻¹⁸ J / 9 = - 2.42x10⁻¹⁹ J

        E₃ =  -13.6 eV / 3² =  -13.6 eV / 9 = - 1.51 eV

The change of energy can be calculated in two ways,

Option 1

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ = 1.93x10⁻¹⁸J

ΔE = E₁ - E₃ = 13.6 - 1.51 = 12.09 eV

Option 2

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE =-13.6 eV (1/nf² - 1/ni²)

Where nf is the final level and ni is the initial level. When the electron passes from its initial level to its final level it is called electronic transition.

• ni = 3
• nf = 1

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE = -2.18x10⁻¹⁸ J (1/1² - 1/3²)

ΔE = -2.18x10⁻¹⁸ J (1 - 0.111)

ΔE = -2.18x10⁻¹⁸ J (0.888)

ΔE = - 1.937x10⁻¹⁸ J

or

ΔE = -13.6 eV  (1/nf² - 1/ni²)

ΔE = -13.6 eV  (1/1² - 1/3²)

ΔE = -13.6 eV  (1 - 0.111)

ΔE = -13.6 eV  (0.888)

ΔE = -12.08 eV

This is the energy required for the electron to go from n= 3 to n = 1. The negative sign (-) means energy (as light or photons) released or emitted.

If we want to express the result in Hz, we just need to make a conversion.

1Hz ⇔ 6.626x10⁻³⁴J ⇔ 4.136x10¹⁵ eV.

The energy required for the electron to go from n= 3 to n = 1 is 3.09x10¹⁵ Hz.

Now, we need to calculate the frequency, ν. This is, how many times the wave oscillates back and foward per second.

To do it, we will use the universal Planck constant, h, and the absolute value of the energy, E.

ν = E/h = 1.937x10⁻¹⁸ J / 6.626x10⁻³⁴ Js = 2.923x10¹⁵ 1/s =  2.923x10¹⁵ Hz.

Answer:

• Frequency, ν = E/h = 2.923x10¹⁵ Hz.
• Energy, E = 3.09x10¹⁵ Hz.

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