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? hints

Answer 1

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

Further explanation:

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  

${{\text{n}}_{\text{1}}}$ 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

Substitute 2 for ${{\text{n}}_{\text{1}}}$ and 5 for ${{\text{n}}_{\text{2}}}$ 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

Substitute 4 for ${{\text{n}}_{\text{1}}}$ and 6 for ${{\text{n}}_{\text{2}}}$ 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

Substitute 2 for ${{\text{n}}_{\text{1}}}$ and 3 for ${{\text{n}}_{\text{2}}}$ 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

Substitute 2 for ${{\text{n}}_{\text{1}}}$ and 6 for ${{\text{n}}_{\text{2}}}$ 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. Ranking of elements according to their first ionization energy: brainly.com/question/1550767
2. Chemical equation representing the first ionization energy for lithium: brainly.com/question/5880605

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Atomic structure

Keywords: Rydberg constant, wavelength, n1, n2, positive integers, transition, 2, 6, 3, 5, transition, Rh, spectral line, shortest wavelength.

Answer 2

n = 6 to n = 2

Further explanation  

From several sources, we have prepared the following answer choices:

A. n = 2 to n = 5  

B. n = 6 to n = 4  

C. n = 3 to n = 2  

D. n = 6 to n = 2  

We will determine which electron transition would correspond to the shortest wavelength line in the visible emission spectra for hydrogen.  

The amount of energy released or absorbed by electrons when moving from n₁ level to n₂ level is equal to  

$\boxed{ \ \Delta E = -13.6 \Big( \frac{1}{n_2^2} - \frac{1}{n_1^2} \Big) \ }$  in eV.

This energy difference is equal to $\boxed{ \ hf = \frac{hc}{\lambda} \ }$, where f and λ are the frequency and wavelength of the radiation emitted or absorbed.

Thus, the wavelength is inversely proportional to the energy difference from the electron transition. To get the shortest wavelength, it is determined by the largest ΔE.  

From the formula above, we practically only need to calculate part $\boxed{ \ \Big( \frac{1}{n_2^2} - \frac{1}{n_1^2} \Big) \ }$ which is directly proportional to ΔE.  Then from the results of the calculation of this section, we will get the shortest wavelength from the largest result..

A. n₁ = 2 to n₂ = 5  

$\boxed{ \ \Big( \frac{1}{5^2} - \frac{1}{2^2} \Big) \ }$  

$\boxed{ \ -\frac{21}{100} \ }$

By taking the absolute value, we get $\boxed{ \ 0.210 \ }$

B. n₁ = 6 to n₂ = 4  

$\boxed{ \ \Big( \frac{1}{4^2} - \frac{1}{6^2} \Big) \ }$  

$\boxed{ \ \frac{5}{144} \ }$

We get $\boxed{ \ 0.0347 \ }$

C. n₁ = 3 to n₂ = 2  

$\boxed{ \ \Big( \frac{1}{2^2} - \frac{1}{3^2} \Big) \ }$  

$\boxed{ \ \frac{5}{36} \ }$

We get $\boxed{ \ 0.1389 \ }$

D. n₁ = 6 to n₂ = 2  

$\boxed{ \ \Big( \frac{1}{2^2} - \frac{1}{6^2} \Big) \ }$  

$\boxed{ \ \frac{2}{9} \ }$

We get $\boxed{ \ 0.222 \ }$

The last calculation above shows the greatest results so that the shortest wavelength is undoubtedly gained from the electron transition n = 6 to n = 2.

Learn more

1. The energy density of the stored energy  brainly.com/question/9617400
2. Particle's speed and direction of motion brainly.com/question/2814900
3. The relationship between a single gold atom and the thickness of the atoms in Rutherford's foil brainly.com/question/4929060  

Keywords: according to the Bohr model of the atom, which electron transition would correspond, to the shortest wavelength, the visible emission spectra for hydrogen, energy, inversely proportional