COMEDK · Physics · 28. Atomic Physics

A hydrogen atom absorbs energy and rises to n = 3 state from its ground state n = 1. If the potential energy of the atom at its ground state is -13.6eV, find the wave length emitted by it when it returns to its ground state:
{Planck's constant= \(6.6 \times 10^{34}\) J s}

A \(7000\)Å
B \(1020\)Å
C \(12000\)Å
D \(4000\)Å

Verified Solution

Answer & Solution

Correct Answer

(B) \(1020\)Å

Step-by-step Solution

Detailed explanation

Total energy of hydrogen atom in ground state, \(E_1 = -13.6\) eV Energy in \(n=3\) state, \(E_3 = \dfrac{-13.6}{3^2} = -1.51\) eV Energy of emitted photon, \(\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\) eV Wavelength of emitted photon, \(\lambda = \dfrac{hc}{\Delta E}\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A hydrogen atom absorbs energy and rises to n = 3 state from its ground state n = 1. If the potential energy of the atom at its ground state is -13.6eV, find the wave length emitted by it when it returns to its ground state:{Planck's constant= \(6.6 \times 10^{34}\) J s}

Step-by-step Solution

A hydrogen atom absorbs energy and rises to n = 3 state from its ground state n = 1. If the potential energy of the atom at its ground state is -13.6eV, find the wave length emitted by it when it returns to its ground state:
{Planck's constant= \(6.6 \times 10^{34}\) J s}