Question

A line in the Brackett series of hydrogen has a wavelength of 1945 nm. From what state did the electron originate?

Answer 1

The electron originated from the 8th shell.

Further Explanation:

An electronic transition is a process that occurs when an electron undergoes emission or absorption from one energy level to another energy level.

When an electron undergoes a transition from a lower energy level to a higher energy level then it requires energy to complete the process. This transition is an absorption process.

When an electron undergoes a transition from higher energy level to lower energy level then it emits energy to complete the process. This transition is an emission process.

The hydrogen atoms are dissociated when an electric discharge is passed through its molecules. As a result, electromagnetic radiations are emitted by the excitation of hydrogen atoms. The hydrogen spectrum contains radiations of different frequencies.

The formula to calculate the wavelength of transition in the hydrogen atom is,

$\dfrac{1}{\lambda }={R_{\text{H}}}\left({\dfrac{1}{{{{\left( {{{\text{n}}_{\text{1}}}}\right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_2}} \right)}^2}}}}\right)$                ......(1)

Here,

$\lambda$ is the wavelength of transition.

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

${{\text{n}}_{\text{1}}}$ is the initial energy level of transition.

${{\text{n}}_{\text{2}}}$ is the final energy level of transition.

When the transition occurs from first energy level to any other level, it is termed as Lyman series.

When the transition occurs from second energy level to any other level, it is termed as Balmer series.

When the transition occurs from third energy level to any other level, it is termed as Paschen series.

When the transition occurs from fourth energy level to any other level, it is termed as Brackett series.

When the transition occurs from fifth energy level to any other level, it is termed as Pfund series

Rearrange equation (1) to calculate ${{\text{n}}_{\text{2}}}$.

$\dfrac{1}{{\left( {{{\text{n}}_2}}\right)}}=\sqrt {\left( {\dfrac{1}{{{{\left( {{{\text{n}}_{\text{1}}}} \right)}^2}}}-\dfrac{1}{{\lambda {R_{\text{H}}}}}} \right)}$

                                             .......(2)

The wavelength is to be converted to m. The conversion factor for this is,

${\text{1 nm}} = {10^{ - 9}}\;{\text{m}}$

So the wavelength can be calculated as follows:

$\begin{aligned}\lambda&= \left({{\text{1945 nm}}} \right)\left( {\frac{{{{10}^{ - 9}}\;{\text{m}}}}{{{\text{1 nm}}}}}\right)\\&= 1945\times {10^{ - 9}}\;{\text{m}}\\\end{aligned}$

The wavelength of Brackett series is $1.945\times {10^{ - 9}}\;{\text{m}}$ .

The initial energy level is 4.

The Rydberg constant is $1.0974\times {10^7}{\text{ }}{{\text{m}}^{ - 1}}$ .

Substitute $1.945\times {10^{ - 9}}\;{\text{m}}$ for $\lambda$ , $1.0974\times {10^7}{\text{ }}{{\text{m}}^{ - 1}}$ for ${R_{\text{H}}}$ and 4 for ${{\text{n}}_{\text{1}}}$ in equation (2).

$\dfrac{1}{{\left( {{{\text{n}}_2}}\right)}}=\sqrt{\left({\dfrac{1}{{{{\left( {\text{4}} \right)}^2}}}-\dfrac{1}{{\left( {1.945 \times {{10}^{ - 9}}\;{\text{m}}}\right)\left( {1.0974 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)}}}\right)}$

Solve for ${{\text{n}}_{\text{2}}}$,

$\begin{aligned}{{\text{n}}_{\text{2}}}&= 7.99377\\&\approx 8\\\end{aligned}$

Therefore the electron originated from the 8th energy level.

Learn more:

1. Which transition is associated with the greatest energy change? brainly.com/question/1594022

2. Describe the spectrum of elemental hydrogen gas: brainly.com/question/6255073

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: hydrogen spectrum, Lyman, Balmer, Paschen, Brackett, Pfund, 8, 1945 nm, 4, RH, n1, n2, first, second, third, fourth, fifth.

Answer 2

Rydberg Eqn is given as:
 1/λ = R [1/n1^2 - 1/n2^2] 
Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m 
R is the Rydberg constant: R = 1.09737×10^7 m-1 
From Brackett series n1 = 4 
Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] 
Some rearranging and collecting up terms: 
1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] 
1= 28.82[1/16 – 1/n2^2] 
28.82/n^2 = 1.8011 – 1 = 0.8011 
n^2 = 28.82/0.8011 = 35.98 
n = √(35.98) = 6