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

Chemistry

1 answer:

never [62]3 years ago

6 0

Answer:

$1.936\times 10^{-18}\ \text{J}$

Explanation:

$R_h$ = Rydberg constant = $2.178\times 10^{-18}\ \text{J}$

$n_i$ = Initial shell = 3

$n_f$ = Final shell = 1

We have the relation

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

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

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..........................................................................................................................

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Hello, I'm attaching a picture with the numerical development of this exercise.

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