Question #03297

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Question #03297

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Answer 1 · 2017-11-21T01:05:36

$n_{f} = 3$ $n_f = 3$

Explanation:

Your tool of choice here will be the Rydberg equation, which allows you to calculate the wavelength of the photon emitted when an electron in a hydrogen atom makes an $n_{i} \to n_{f}$ $n_i -> n_f$ transition.

$\frac{1}{λ} = R \cdot (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ $1/(lamda) = R * (1/n_f^2 - 1/n_i^2)$

Here

$R$ $R$ is the Rydberg constant, equal to $1.097 \cdot 10^{7}$ $1.097 * 10^(7)$ $m^{- 1}$ $"m"^(-1)$

$n_{i}$ $n_i$ is the energy level from which the electron falls

$n_{f}$ $n_f$ is the energy level to which the electron falls

In your case, you know that the electron falls from the seventh energy level, so

$n_{i} = 7$ $n_i = 7$

Your goal here is to find the value of the final energy level, $n_{f}$ $n_f$ . Rearrange the Rydberg equation to solve for $n_{f}$ $n_f$ .

$\frac{1}{λ} = R \cdot \frac{n_{i}^{2} - n_{f}^{2}}{n_{i}^{2} \cdot n_{f}^{2}}$ $1/lamda = R * (n_i^2 - n_f^2)/(n_i^2 * n_f^2)$

$n_{i}^{2} \cdot n_{f}^{2} = λ \cdot R \cdot (n_{i}^{2} - n_{f}^{2})$ $n_i^2 * n_f^2 = lamda * R * (n_i^2 - n_f^2)$

$n_{i}^{2} \cdot n_{f}^{2} + λ \cdot R \cdot n_{f}^{2} = λ \cdot R \cdot n_{i}^{2}$ $n_i^2 * n_f^2 + lamda * R * n_f^2 = lamda * R * n_i^2$

This is equivalent to

$n_{f}^{2} \cdot (n_{i}^{2} + λ \cdot R) = λ \cdot R \cdot n_{i}^{2}$ $n_f^2 * (n_i^2 + lamda * R) = lamda * R * n_i^2$

which gets you

$n_{f} = \sqrt{\frac{n_{i}^{2} \cdot λ \cdot R}{n_{i}^{2} + λ \cdot R}}$ $n_f = sqrt( (n_i^2 * lamda * R)/(n_i^2 + lamda * R))$

Now all you have to do is to plug in the values that you have--do not forget that the wavelength of the photon must be expressed in meters!

$n_f = sqrt( (7^2 * 1005 * 10^(-9)color(red)(cancel(color(black)("m"))) * 1.097 * 10^7 color(red)(cancel(color(black)("m"^(-1)))))/(7^2 + 1005 * 10^(-9)color(red)(cancel(color(black)("m"))) * 10.97 * 10^7 color(red)(cancel(color(black)("m"^(-1))))))$

$n_f = 2.99983 ~~ 3$

Therefore, you can say that a photon of wavelength $"1005 nM"$ is emitted when an electron in a hydrogen atom falls from $n_i = 7$ to $n_f = 3$ , a transition that is part of the Paschen series.

![https://thecuriousastronomer.wordpress.com/2013/08/20/emission-line-spectra/]( useruploads.socratic.org )

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Question #03297

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