When a hydrogen atom is bombared, the atom is excited to then $n = 4$ state. The energy released, when the atom goes from $n = 4$ state to the ground state is

Question

Dr. Abubacker · Accepted Answer

(b) Energy released $= E_{4} - E_{1}$
$= - \frac{13.6}{4^{2}} - \left( - \frac{13.6}{1^{2}} \right) = 1.75eV$

Q. When a hydrogen atom is bombared, the atom is excited to then $n = 4$ state. The energy released, when the atom goes from $n = 4$ state to the ground state is

Explanation

Reviewed by Dr. Abubacker

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