How would you calculate the energy released as a photon drops from n=3 to the ground state?

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How would you calculate the energy released as a photon drops from n=3 to the ground state?

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Answer 1 · 2017-05-13T21:24:16

$DeltaE$ = $-1.936×10^-18J$

Explanation:

$DeltaE$ = $K$ $(1/n_1^2-1/n_2^2)$
Where $n_1$ = $1$ and $n_2$ =3, since the photon falls from excited to ground state the energy is released.

$DeltaE$ = $K$ $(1/1^2-1/3^2)$
$DeltaE$ = $K$ $(1/1-1/9)$
But $K$ = $-2.178×10^-18J$ , a constant value for calculating energy of an electron in H atom
$DeltaE$ = $-2.178×10^-18$ $(1/1-1/9)$
$DeltaE$ = $-1.936×10^_18J$
The answer is for H-atom.You didn't mentioned about the atom for which you are asking for?

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How would you calculate the energy released as a photon drops from n=3 to the ground state?

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