Question:medium

The energy of hydrogen atom in ground state \( (n = 1) \) is \( -13.6 \) eV. To which highest energy-state the hydrogen atom can be made to reach by the photon of energy \( 12.09 \) eV?
OR
What is meant by interference? Write the conditions for constructive and destructive interference.

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For the atom, add the photon energy to the ground-state energy and match it to \( -13.6/n^2 \). For interference, recall the path-difference conditions \( n\lambda \) and \( (2n-1)\lambda/2 \).
Updated On: Jul 10, 2026
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Solution and Explanation

Option 1 (Atoms):
Step 1: Read the absorbed photon as an upward transition from the ground level \(n_i = 1\) to a higher level \(n_f = n\). The photon energy equals the energy gap: \(E_{photon} = 13.6\left(\dfrac{1}{n_i^2} - \dfrac{1}{n_f^2}\right)\) eV.
Step 2: Substitute: \(12.09 = 13.6\left(1 - \dfrac{1}{n^2}\right)\).
Step 3: \(1 - \dfrac{1}{n^2} = \dfrac{12.09}{13.6} = 0.889\), so \(\dfrac{1}{n^2} = 0.111\).
Step 4: \(n^2 = \dfrac{1}{0.111} = 9.0 \Rightarrow n = 3\). The highest reachable level is the third one.
\[\boxed{n = 3}\]

Option 2 (Interference):
Step 1: When two coherent waves \(y_1 = a\sin\omega t\) and \(y_2 = a\sin(\omega t + \phi)\) meet, their amplitudes combine to give a resultant amplitude \(R = 2a\cos(\phi/2)\), and the intensity follows \(I \propto R^2\).
Step 2: Maximum intensity (constructive) needs \(\cos(\phi/2) = \pm 1\), i.e. phase difference \(\phi = 2n\pi\), which is a path difference of \(n\lambda\).
Step 3: Minimum intensity (destructive) needs \(\cos(\phi/2) = 0\), i.e. \(\phi = (2n-1)\pi\), which is a path difference of \((2n-1)\lambda/2\).
\[\boxed{\text{Bright: path diff } n\lambda;\ \text{Dark: path diff } (2n-1)\lambda/2}\]
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