Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Complete Question
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm. Group of answer choices
Answer:
[tex]n=2[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=486nm=>486*10^{-9}[/tex]
Generally the equation for Atom Transition is mathematically given by
[tex]\frac{1}{\lambda}=R_{\infty }(\frac{1}{n_1^2}-\frac{1}{n_2^2})[/tex]
Where
Rydberg constant [tex]R_{\infty}=1.097*10^7[/tex]
Therefore
[tex]\frac{1}{486*10^{-9}}=1.097*10^7*(\frac{1}{n_1^2}-\frac{1}{4^2})[/tex]
[tex](\frac{1}{n_1^2}-\frac{1}{4^2})=\frac{1}{486*10^{-9}*1.097*10^7}[/tex]
[tex]n_1^2=3.98[/tex]
[tex]n=1.99[/tex]
[tex]n=2[/tex]
Using the Rydberg formula, the final state of the electron is n=2.
Using the Rydberg formula;
1/λ = R(1/nf^2 - 1/ni^2)
Where;
λ = wavelength
nf = final state
ni = initial state
R = Rydberg constant
When λ = 486 × 10^-9 m and ni = 4, R = 1.097 × 10^7 m-1
1/486 × 10^-9 = 1.097 × 10^7(1/nf^2 - 1/4^2)
0.188 = 1/nf^2 - 0.0625
1/nf^2 = 0.188 + 0.0625
nf = 2
Missing parts;
Determine the end (final) value of n in the hydrogen atom transition, if electron starts in n-4 and the atom emits a photon of light with a wavelength of 486.
Learn more: https://brainly.com/question/14281129
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.