Chemistry, 21.11.2019 11:31 twistedhyperboles

Given the energy levels of electrons in a hydrogen atom (table), rank the photons emitted by the following electron transitions from longest to shortest wavelength.

n=4 to n=3
n=4 to n=2
n=2 to n=1
n=3 to n=1

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Answer from: zahraa244

Longest to shortest wavelength:

n=4 to n=3 > n = 4 to n= 2 > n = 2 to n= 1 > n = 3 to n= 1

Explanation:

The energy levels in a hydrogen atom vary based on the formula:

$E_{n} = \frac{13.6 eV}{n^{2}}$

where 'n' corresponds to the energy levels associated with a transition and can take on integer values 1,2,3...

Therefore, based on the above relation, the energy separation decreases rapidly as n increases. Since energy and the wavelength of a photon are inversely related, lower the energy larger will be the wavelength and vice versa.

The wavelength of photons from the longest (lowest energy, smallest separation between energy levels) to the shortest (highest energy, large separation between energy levels) would be:

n=4 to n=3 > n = 4 to n= 2 > n = 2 to n= 1 > n = 3 to n= 1

Answer from: andybiersack154

The order of wavelength $\left(\lambda\right)$ of the emitted photons from longest to shortest is,

$\boxed{{\lambda _1}\left( {4 \to 3}\right){\lambda _2}\left( {4\to2}\right){\lambda _3}\left( {2\to1}\right){\lambda _4}\left( {3\to1}\right)}$

Further explanation:

Concept:

According to the equation of photon energy, the energy of the photon is inversely proportional to the wavelength.

$E=\dfrac{{hc}}{\lambda }$

Here, E is the photon energy, h is the Planck constant, $\left(\lambda\right)$ is the wavelength of photon, and c is the speed of light.

The formula to calculate the energy of the emitted photon in the hydrogen atom is,

$E=\left({-13.6{\text{ eV}}}\right)\left({\dfrac{1}{{{{\left({{{\text{n}}_{\text{i}}}} \right)}^2}}}-\dfrac{1}{{{{\left( {{{\text{n}}_{\text{f}}}}\right)}^2}}}}\right)$

Where,

E is the energy of emitted photon.

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

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

Solution:

Finding the energy of emitted photons in each transition.

1. For the first transition, from initial energy level n=4 to final energy level n=3.

$\begin{aligned}{E_1}&=\left({-13.6{\text{ eV}}}\right)\left({\dfrac{1}{{{{\left({{{\text{n}}_{\text{i}}}} \right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}} \right)}^2}}}}\right)\\&=\left( { - 13.6{\text{ eV}}} \right)\left( {\frac{1}{{{{\left( {\text{4}}\right)}^2}}}-\dfrac{1}{{{{\left( {\text{3}}\right)}^2}}}}\right)\\&=0.6611\,{\text{eV}}\\\end{aligned}$

2. For the second transition, from initial energy level n=4 to final energy level n=2.

$\begin{aligned}{E_1}&=\left({-13.6{\text{ eV}}}\right)\left({\dfrac{1}{{{{\left({{{\text{n}}_{\text{i}}}} \right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}} \right)}^2}}}}\right)\\&=\left( { - 13.6{\text{ eV}}} \right)\left( {\frac{1}{{{{\left( {\text{4}}\right)}^2}}}-\dfrac{1}{{{{\left( {\text{2}}\right)}^2}}}}\right)\\&=2.55\,{\text{eV}}\\\end{aligned}$

3. For the third transition, from initial energy level n=2 to final energy level n=1.

$\begin{aligned}{E_1}&=\left({-13.6{\text{ eV}}}\right)\left({\dfrac{1}{{{{\left({{{\text{n}}_{\text{i}}}} \right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}} \right)}^2}}}}\right)\\&=\left( { - 13.6{\text{ eV}}} \right)\left( {\frac{1}{{{{\left( {\text{2}}\right)}^2}}}-\dfrac{1}{{{{\left( {\text{1}}\right)}^2}}}}\right)\\&=10.2\,{\text{eV}}\\\end{aligned}$

4. For the fourth transition, from initial energy level n=3 to final energy level n=1.

$\begin{aligned}{E_1}&=\left({-13.6{\text{ eV}}}\right)\left({\dfrac{1}{{{{\left({{{\text{n}}_{\text{i}}}} \right)}^2}}}-\frac{1}{{{{\left({{{\text{n}}_{\text{f}}}} \right)}^2}}}}\right)\\&=\left( { - 13.6{\text{ eV}}} \right)\left( {\frac{1}{{{{\left( {\text{3}}\right)}^2}}}-\dfrac{1}{{{{\left( {\text{1}}\right)}^2}}}}\right)\\&=12.09\,{\text{eV}}\\\end{aligned}$

The energy of emitted photons in increasing order are,

${E_1}\left({4 \to 3} \right)$

Since the wavelength is inversely proportional to the energy of emitted photons, therefore, corresponding order of wavelength $\left(\lambda\right)$ of the emitted photons from longest to shortest is,

$\boxed{{\lambda _1}\left( {4 \to 3}\right){\lambda _2}\left( {4\to2}\right){\lambda _3}\left( {2\to1}\right){\lambda _4}\left( {3\to1}\right)}$

Learn more:

1. Ranking of elements according to their first ionization energy.:

2.Chemical equation representing the first ionization energy for lithium.:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: transition, hydrogen atom, energy difference, transition from n=4 to n=3, transition from n=4 to n=2, transition from n=2 to n=1, transition from n=3 to n=1, greatest, emitted photons.

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