Respuesta :
Answer: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:
[tex]l=\frac{(n_2-n_1)(n_2-n_1+1)}{2}[/tex]
[tex]n_1[/tex]=initial energy level
[tex]n_2[/tex] = final energy level.
Hydrogen atoms are excited by a laser to the n = 4 state .So the number of spectral lines observed will be:
[tex]l=\frac{(4-1)(4-1+1)}{2}=6[/tex]
The wavelength of the 4 to 2 transition will be given by Rydberg expression:
[tex]\frac{1}{\lambda }=R[\frac{1}{n_{1}^2}-\frac{1}{n_{2}^2}][/tex]
[tex]\frac{1}{\lambda }=1.097\times 10^{10}\times [\frac{1}{4^2}-\frac{1}{2^2}][/tex]
[tex]\lambda =486.17 nm[/tex] ([tex]1nm =10^{-9} m[/tex])
The maximum number spectral lines observed will be 6.
The wavelength of the 4 to 2 transition will be 486.17 nm.
The total number of spectral line for hydrogen atom at n=4 is [tex]\boxed6[/tex].
The value of wavelength of a spectral line of transition from n=4 to n=2 is[tex]\boxed{{\text{486}}{\text{.2 nm}}}[/tex].
Further explanation:
The total number of spectral line from a given initial energy level is,
[tex]{\text{Total number of spectral line}}=\frac{{{\text{n}}\left({{\text{n}}-1}\right)}}{2}[/tex] …… (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).
[tex]\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}[/tex]
According to the Rydberg equation, the wavelength of spectral line related with the transition values as follows:
[tex]\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)[/tex] …… (1)
Here, is the wavelength of spectral line, [tex]{{\text{R}}_{\text{H}}}[/tex] is the Rydberg constant that has the value[tex]1.097\times{10^7}{\text{}}{{\text{m}}^{-1}}[/tex], [tex]{{\text{n}}_{\text{i}}}[/tex] is the initial energy level of transition, and [tex]{{\text{n}}_{\text{f}}}[/tex] is the final energy level of transition.
Therefore, after rearrangement of equation (1) can be calculated as,
[tex]\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)}}[/tex] …… (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
[tex]\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}[/tex]
Learn more:
1. Ranking of elements according to their first ionization energy.:https://brainly.com/question/1550767
2. Chemical equation representing the first ionization energy for lithium.:https://brainly.com/question/5880605
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.
