Respuesta :
Answer:
(a) n₂ = 2
(b) n₂ = 5
(c) n₂ = 4
(d) n₂ = 3
(e) n₂ = 6
Explanation:
The Rydberg equation give us the wavelength of the transition between energy levels according to the formula:
1/λ = Rh x ( 1/n₁² - 1/n₂² )
where n₁ is the final state and n₂ is the initial state.
The strategy here, since we are given the wavelength, is to solve for λ, and then by substituting for n₂ combinations find which ones match our question.
λ = 1 / [ Rh x ( 1/n₁² - 1/n₂² ) ]
Lets express Rh in 1/Angstrom
1.097 x 10 ⁷ / [m x ( 1 m/ 10¹⁰ A) ] = 0.011 / Å
⇒ λ = 1 / [0.011 A x ( 1/n₁² - 1/n₂² )] = 911.6 Å / ( 1/n₁² - 1/n₂² )
For n₁ = 2 n₂ = 3, 4, 5,.......
λ ( n₂ = 3 ) = 911.6 A / ( 1/2² - 1/3² ) = 6563.5 Å
λ ( n₂ = 4 ) = 911.6 A / ( 1/2² - 1/4² ) = 4861.9 Å
λ ( n₂ = 5 ) = 911.6 A / ( 1/2² - 1/5² ) = 4341.0 Å
So we have matched three of the transitions
Now for n₁ = 1 n₂ = 2, 3, 4....
λ ( n₂ = 2 ) = 911.6 A / ( 1/1² - 1/2² ) = 1215.5 Å
For n₁ = 3 n₂ = 4, 5, 6....
λ ( n₂ = 4 ) = 911.6 A / ( 1/3² - 1/4² ) = 18752.9 Å
λ ( n₂ = 5 ) = 911.6 A / ( 1/3² - 1/5² ) = 12819.4 Å
λ ( n₂ = 6 ) = 911.6 A / ( 1/3² - 1/6² ) = 10939.2 Å
a. For [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.
b. For [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.
c. For [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.
d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.
e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.
The wavelength from the transition has been given by the Rydberg equation as:
[tex]\dfrac{1}{\lambda}=\text Rh\dfrac{1}{N_1}-\dfrac{1}{N_2}[/tex]
Where, wavelength of the radiation, [tex]\lambda[/tex]
The initial transition level, [tex]N_1[/tex]
The final transition level, [tex]N_2[/tex]
The constant [tex]Rh=1.097\;\times\;10^7[/tex], or in Armstrong it can be given as, [tex]0.011/\r{\rm A}[/tex]
The wavelength can be given as:
[tex]\lambda=\dfrac{1}{0.011\;\r{\rm A}\;\times\;(\frac{1}{N_1^2} -\frac{1}{N_2^2}) }[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{N_1^2}-\frac{1}{N_2^2} }[/tex]
a. The initial level ([tex]N_2[/tex]) of transition has been given for [tex]\lambda=1212.7\;\rm \r{A}[/tex], and [tex]N_2=2[/tex] has been given, with substituting [tex]N_1=1[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{1^2}-\frac{1}{2^2} }\\\lambda=1215.5\;\r{A}[/tex]
Thus, for [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.
b. For, [tex]\lambda=4340.5\;\rm \r{A}[/tex], and [tex]N_2=2[/tex], [tex]N_1=5[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{5^2} }\\\lambda=4341.0\;\r{A}[/tex]
Thus, for [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.
c. For, [tex]\lambda=4861.3\;\rm \r{A}[/tex], the final transition has been, [tex]N_2=2[/tex], the initial level has been substituted as [tex]N_1=4[/tex]:
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{4^2} }\\\lambda=4861.3\;\r{A}[/tex]
Thus, for [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.
d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], the final level has been 2, the initial level has been substituted as, [tex]N_1=3[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{3^2} }\\\lambda=6562.8\;\r{A}[/tex]
Thus, for [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.
e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level has been substituted as, [tex]N_1=6[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{3^2}-\frac{1}{6^2} }\\\lambda=10,938\;\r{A}[/tex]
Thus, for [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.
For more information about transition levels, refer to the link:
https://brainly.com/question/4511550
