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Calculate the wavelength of the radiation released when an electron in a hydrogen atom moves from $n = 5$ to $n = 2$.

How can I find this wavelength?

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The Rydberg formula (for the hydrogen atom) is usually given as

$$\tilde{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_\mathrm f^2}-\frac{1}{n_\mathrm i^2} \right)$$

Here, $R$ is the Rydberg constant; take care of the unit ($\mathbf{m^{-1}}$ or $\mathbf{cm^{-1}}$) here! Wikipedia quotes a value of $\pu{1.097 \times 10^7 m-1}$ for a nucleus of infinite mass. In principle, the value for a nucleus of finite mass (as is the case here) is slightly different, but for the purposes of this question, we can ignore that.

$n_\mathrm f$ is the final state, and $n_\mathrm i$ is the initial state.

So, if you substitute $R = \pu{1.097 \times 10^7 m-1}$, $n_\mathrm{f} = 2$, $n_\mathrm i = 5$, you can arrive at $$\tilde\nu = \pu{2.304 \times 10^6 m-1}$$

and the wavelength is $\lambda = 1/\tilde\nu = \pu{4.341 \times 10^-7 m}$, or if you convert to nanometers, $\lambda = \pu{434.1 nm}$. Double-checking against the Wikipedia page on the Balmer series, this value is indeed correct.

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Start by using Rydberg formula ν~=1/λ=R(1/n2f−1/n2i) =1.10x10^7 m^-1 (1/2^2 - 1/5^2) = 2.31x10^6 this positive value means it is absorbing that energy instead of releasing it in the form of a photon.

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    $\begingroup$ Welcome to Chem SE! This answer looks reasonable, but you should edit the equations using MathJax (explained here) $\endgroup$ – Tyberius Aug 29 '17 at 14:39

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