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lawyer [7]
3 years ago
8

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

n returns to the ground state by emitting two photons. What are their wavelengths?
Physics
1 answer:
IRISSAK [1]3 years ago
7 0

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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Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

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The buoyant force is equal to the weight of displaced water, and the
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The weight of the displaced water is 30N, and weight = (mass) (gravity).

           30N = (mass of the displaced water) x (9.81 m/s²)

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================================================

I'm thinking that this must  be the hard way to do it,
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So apparently . . .

        (density of a sample) / (density of water) =

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I never knew that, but it's a good factoid to keep in my tool-box.


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