Question

According to the Bohr model of the atom, which electron transition would correspond to the shortest wavelength line in the visible emission spectra for hydrogen?

Answer 1

Explanation:

According to Bohr's model, electrons are travelled in defined circular orbit around the nucleus. The orbits of different energy level is labelled by quantum numbers (n). Electrons can jump different energy levels, from higher to lower energy level by releasing energy.

In the visible emission spectrum of hydrogen, there are mainly five series which we are studied in regular. Series are: Lyman series(n=1), Balmer series(n=2), Paschen series(n=3), Bracket series(n=4), Pfund series(n=5).

Wavelength is inversely proportional to the quantum number(n). So, we can conclude that greater the value of n shortest the wavelength. According to above spectral series Pfund series(n=5) have shortest wavelength.

 

Answer 2

The electronic transition from $\boxed{{\text{D}}{\text{. n}} = {\text{6 to n}} = {\text{2}}}$ corresponds to the shortest wavelength.

Further explanation:

Electronic transition is a process that takes place when an electron undergoes either emission or absorption from one energy level to another energy level.

When transition of electrons occurs from lower to higher energy level, energy is needed for the process and such transition is an absorption process.

When an electron experiences a transition from higher to lower energy level, it emits energy to complete the process and such transition is an emission process.

Rydberg equation describes the relation of wavelength of spectral line with the transition values. The expression for Rydberg equation is as follows:

$\dfrac{1}{\lambda }= \left( {{{\text{R}}_{\text{H}}}} \right)\left( {\dfrac{1}{{{{\left( {{{\text{n}}_{\text{1}}}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)$                                                     …… (1)

Here,

$\lambda$ is the wavelength of spectral line

${{\text{R}}_{\text{H}}}$ is Rydberg constant that has the value  $1.097 \times {10^7}{\text{ }}{{\text{m}}^{ - 1}}$

${{\text{n}}_{\text{1}}}$ and ${{\text{n}}_{\text{2}}}$ are the two positive integers, where  .

Rearrange equation (1) to calculate  .

$\lambda = \Dfrac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\Dfrac{1}{{{{\left( {{{\text{n}}_1}}\right)}^2}}} - \Dfrac{1}{{{{\left({{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)}}$                                        …… (2)

A. n = 2 to n = 5

The value of ${{\text{n}}_{\text{1}}}$ is 2.

The value of ${{\text{n}}_{\text{2}}}$ is 5.

Substitute these values in equation (2).

$\begin{aligned}\lambda &= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left({\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 5 \right)}^2}}}}\right)}} \\&= 4.34 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}$  

B. n = 6 to n = 4

The value of ${{\text{n}}_{\text{1}}}$ is 4.

The value of ${{\text{n}}_{\text{2}}}$ is 6.

Substitute these values in equation (2).

$\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( 4 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}} \right)}}\\&= 2.63 \times {10^{ - 6}}{\text{ m}}\\\end{aligned}$

C. n = 3 to n = 2

The value of ${{\text{n}}_{\text{1}}}$ is 2.

The value of ${{\text{n}}_{\text{2}}}$ is 3.

Substitute these values in equation (2).

 $\begin{aligned}\lambda &= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 3 \right)}^2}}}} \right)}} \\&= 6.56 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}$

D. n = 6 to n = 2

The value of ${{\text{n}}_{\text{1}}}$ is 2.

The value of ${{\text{n}}_{\text{2}}}$ is 6.

Substitute these values in equation (2).

$\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left( {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}} \right)}} \\ &= 4.10 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}$

The value of $\lambda$ for transition from n = 6 to n = 2 is the least and therefore this transition corresponds to the shortest wavelength.

Learn more:

1. Rank the transitions from longest to shortest wavelength: https://brainly.com/question/2055545
2. Calculate the wavelengths of spectral transitions of hydrogen atom: https://brainly.com/question/1857156

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: Rydberg constant, wavelength, positive integers, n1, n2, transition, Rh, spectral line, shortest wavelength, transition.