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Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =3 to =1.

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Answer:

[tex]1.936\times 10^{-18}\ \text{J}[/tex]

Explanation:

[tex]R_h[/tex] = Rydberg constant = [tex]2.178\times 10^{-18}\ \text{J}[/tex]

[tex]n_i[/tex] = Initial shell = 3

[tex]n_f[/tex] = Final shell = 1

We have the relation

[tex]\Delta E=R_h(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})\\\Rightarrow \Delta E=2.178\times 10^{-18}(\dfrac{1}{1^2}-\dfrac{1}{3^2})\\\Rightarrow \Delta E=1.936\times 10^{-18}\ \text{J}[/tex]

The energy of the photon emitted here is [tex]1.936\times 10^{-18}\ \text{J}[/tex].

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