Chemistry, 17.07.2019 06:28, quita03

Hydrogen atoms are excited by a laser to the n = 4 state and then allowed to emit. what is the maximum number of distinct emission spectral lines (lines of different wavelengths) that can be observed from this system? calculate the wavelength of the 4 to 2 transition.

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answered: Miguel9825

The maximum number spectral lines observed will be 6.

The wavelength of the 4 to 2 transition will be 486.17 nm.

Explanation:

The formula fro number of spectral lines are given by:

$l=\frac{(n_2-n_1)(n_2-n_1+1)}{2}$

n_1 =initial energy level

n_2 = final energy level.

Hydrogen atoms are excited by a laser to the n = 4 state .So the number of spectral lines observed will be:

$l=\frac{(4-1)(4-1+1)}{2}=6$

The wavelength of the 4 to 2 transition will be given by Rydberg expression:

$\frac{1}{\lambda }=R[\frac{1}{n_{1}^2}-\frac{1}{n_{2}^2}]$

$\frac{1}{\lambda }=1.097\times 10^{10}\times [\frac{1}{4^2}-\frac{1}{2^2}]$

$\lambda =486.17 nm$ ( $1nm =10^{-9} m$ )

The maximum number spectral lines observed will be 6.

The wavelength of the 4 to 2 transition will be 486.17 nm.

answered: Kelseyzim

The total number of spectral line for hydrogen atom at n=4 is $\boxed6$ .

The value of wavelength of a spectral line of transition from n=4 to n=2 is $\boxed{{\text{486}}{\text{.2 nm}}}$ .

Further explanation:

The total number of spectral line from a given initial energy level is,

${\text{Total number of spectral line}}=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}$ …… (1)

Here, n is the initial energy level of transition.

In the given question, the initial energy level of hydrogen atom is at 4, so substitute 4 for n in equation (1).

$\begin{aligned}{\text{Total number of spectral line}}&=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}\\&=\frac{{4\left({4-1}\right)}}{2}\\&=6\\\end{aligned}$

According to the Rydberg equation, the wavelength of spectral line related with the transition values as follows:

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

Here, is the wavelength of spectral line, ${{\text{R}}_{\text{H}}}$ is the Rydberg constant that has the value $1.097\times{10^7}{\text{}}{{\text{m}}^{-1}}$ , ${{\text{n}}_{\text{i}}}$ is the initial energy level of transition, and ${{\text{n}}_{\text{f}}}$ is the final energy level of transition.

Therefore, after rearrangement of equation (1) can be calculated as,

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

Finding the wavelength of spectral lines in transition from n=4 to n=2

The initial energy level n=4 to final energy level n=2. Substitute the values of

$\begin{aligned}{\lambda _3}&=\frac{1}{{\left({1.097\times{{10}^7}{\text{ }}{{\text{m}}^{-1}}}\right)\left({\frac{1}{{{{\left({\text{2}}\right)}^2}}}-\frac{1}{{{{\left({\text{4}}\right)}^2}}}}\right)}}\\&=4.862\times{10^{-7}}{\text{ m}}\times\left({\frac{{1{\text{ nm}}}}{{{{10}^{-9}}{\text{ m}}}}}\right)\\&=486.2{\text{ nm}}\\\end{aligned}$

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=2, spectral lines, wavelength of spectral lines, and 486.2 nm.

answered: Guest

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answered: Guest

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