Question

For a hydrogen- like atom, classify these electron transitions by whether they result in the absorption or emission of light: n=3 to n=5, n=1 to n=3, n=3 to n=2, n=2 to n=1? ignoring the sign, which transition was associated with the greatest energy change

Answer 1

The electron transfer from n = 1 to n = 3 occurs the greatest energy change

Further explanation

In an atom, there are energy levels in the skin and sub skin.

This energy level is expressed in terms of electron configurations.

Writing the electron configuration starts from the lowest to the highest subshell's energy level. There are 4 sub-shells in an atom's shell, namely s, p, d, and f. The maximum number of electrons for each subshell is

• s: 2 electrons

• p: 6 electrons

• d: 10 electrons and

• f: 14 electrons

The filling of electrons uses the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Each sub-shell also has orbitals drawn in the form of a square box in which there are electrons symbolized by half arrows.

Each orbital in an atom consists of 4 quantum numbers, namely the main quantum number (n), the azimuth quantum number (l), and the magnetic quantum number (m) and the spin quantum number (s)

• Value of n: positive integer

• value of l: between 0 to n-1

• m value: between -l to + l

• value s: +1/2 or -1/2

Determination of electron configurations based on principles:

• 1. Aufbau: Electrons occupy orbitals of the lowest energy level

• 2 Hund: electron fills orbitals with the same energy level

• 3. Pauli: there are no electrons that have 4 equal quantum numbers

Bohr's atomic model has shown the energy levels of electrons in the path of the atomic shell

The greater the value of n (the atomic shell, the main quantum number), the greater the energy level

In normal circumstances, electrons fill the skin at the lowest energy level starting from the skin K, L M and then N

When an atom gets energy from outside, the electrons will absorb energy so it moves to higher energy. This situation is called excited

Electrons will return to the original path or a lower energy level because the excited state is unstable. In this condition, the electron will release energy

The electron energy at the nth path can be formulated:

En = -Rh / n²

Rh = constant 2.179.10⁻¹⁸ J

So the electron transfer energy (delta E)

delta E = E end - E start

From the electron transfer available, because the value of the Rh constant is the same, the effect is the value of n (skin) ---> 1/n²

• 1. n = 3 to n = 5

then the value of E is

delta E = -1/25 +1/9 = 0.07

• 2. n = 1 to n = 3,

then the value of E is

delta E = -1/9 + 1/1 = 0.89

• 3.n = 3 to n = 2,

then the value of E is

delta E = -1 / 9 + 1/16 = 0.14

• 4. n = 2 to n = 1

then the value of E is

delta E = -1/4 + 1 = 0.75

These 4 values ​​indicate that regardless of the sign that there is an electron transfer from n = 1 to n = 3 the greatest energy change occurs

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Keywords: electron transfer, quantum number, electron shell

Answer 2

Absorption occurs when electron transition from n = 3 to n = 5 energy level and n = 1 to n = 3 energy level.

Emission occurs when electron transition from n = 3 to n = 2 energy level and n = 2 to n = 1 energy level.

The greatest energy change is observed in transition from n=1 energy level to n=3 energy level.

Further explanation:

Emission or absorption of energy by electron while going from one energy level to another level is described in terms of electronic transition.

Absorption takes place when electron goes from lower to higher energy level while emission occurs when electron jumps from higher to lower energy levels.

The formula to calculate the difference between two energy levels of the hydrogen atom is,

$\Delta E=R_{\text{H}}\left(\dfrac{1}{\text{n}^{2}_i}-\dfrac{1}{\text{n}^{2}_f}\right)$                                      ...... (1)

Where,

$\Delta E$ is the energy difference between two energy levels.

$R_{\text{H}}$ is a Rydberg constant.

$\text{n}_i$ is the initial energy level of transition.

$\text{n}_f$is the final energy level of transition .

Since transition occurs from lower to higher energy level (3 to 5), it is an absorption process.

Since transition occurs from lower to higher energy level (1 to 3), it is an absorption process.

Since transition occurs from higher to lower energy level (3 to 2), it is an absorption process.

Since transition occurs from higher to lower energy level (2 to 1), it is an absorption process.

Substitute 3 for  $\text{n}_i$ and 5 for  $\text{n}_f$ in equation (1).

$\begin{aligned}\Delta E&=R_{\text{H}}\left(\dfrac{1}{\text{3}^{2}}-\dfrac{1}{\text{5}^{2}}\right)\\&=0.071R_{\text{H}}\end{aligned}$

Substitute 1 for  $\text{n}_i$ and 3 for  $\text{n}_f$ in equation (1).

$\begin{aligned}\Delta E&=R_{\text{H}}\left(\dfrac{1}{\text{1}^{2}}-\dfrac{1}{\text{3}^{2}}\right)\\&=0.88R_{\text{H}}\end{aligned}$

Substitute 3 for  $\text{n}_i$ and 2 for  $\text{n}_f$ in equation (1).

$\begin{aligned}\Delta E&=R_{\text{H}}\left(\dfrac{1}{\text{3}^{2}}-\dfrac{1}{\text{2}^{2}}\right)\\&=-0.139R_{\text{H}}\end{aligned}$

Substitute 2 for  $\text{n}_i$ and 1 for  $\text{n}_f$ in equation (1).

$\begin{aligned}\Delta E&=R_{\text{H}}\left(\dfrac{1}{\text{2}^{2}}-\dfrac{1}{\text{1}^{2}}\right)\\&=-0.75R_{\text{H}}\end{aligned}$

The energy difference is highest in transition from first energy level to third energy level so it is associated with the highest energy difference.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: Transition, emission, absorption, lower, higher, energy difference, ni, nf, initial level, final level, Rydberg constant, electronic transition.