A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to

Question

A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to the ground state by emitting two photons. What are their wavelengths?

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Maris 4 years 2021-09-05T03:37:51+00:00 1 Answers 9 views 0

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  1. quangkhai
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    2021-09-05T03:39:40+00:00
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    Answer:

    \lambda=25.6nm

    Explanation:

    The Rydberg formula can be extended for use with any hydrogen-like chemical elements, that is to say with only one electron being affected by effective nuclear charge. So, in this case, we can calculate the wavelenghts of the emitted photons using this formula:

    \frac{1}{\lambda}=RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})

    Where R is the Rydberg constant of the element, Z its atomic number, n_1 is the lower energy level and n_2 the upper energy level of the  electron transition. Recall that the ground state is denoted as n=1.

    \frac{1}{\lambda}=1.1*10^7m^{-1}(2)2^2(\frac{1}{1^2}-\frac{1}{3^2})\\\frac{1}{\lambda}=3.91*10^7m^{-1}\\\lambda=2.56*10^{-8}m=25.6nm

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