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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.
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