\(ΔE=\frac{hc}{λ} ⇒ λ\alpha \frac{1}{ΔE}\) For shortest wavelength, energy gap should be maximum. So, correct choice is transition from n = 3 to n = 1.
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Approach Solution -2
Photon Emission and Wavelength Problem
Step 1: Energy and Wavelength Relationship
The energy of a photon emitted during a transition is related to its wavelength by:
\( E = \frac{hc}{\lambda} \)
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
Step 2: Shortest Wavelength Condition
From the equation above, we can see that for the shortest wavelength (\( \lambda_{min} \)), the energy (\( E \)) must be maximum.
Step 3: Energy Difference and Transition
The energy of a photon emitted during a transition is equal to the difference in energy levels:
\( E = E_{initial} - E_{final} \)
The largest energy difference corresponds to the shortest wavelength. In the given diagram:
Transition A: \( n = 4 \) to \( n = 3 \)
Transition B: \( n = 4 \) to \( n = 2 \)
Transition C: \( n = 3 \) to \( n = 1 \)
Transition D: \( n = 3 \) to \( n = 1 \) (Likely an error in the original question)
The energy levels in a hydrogen atom are given by:
\( E_n = -\frac{13.6 \text{ eV}}{n^2} \)
We can see that transitions C and D are identical in the provided image and diagram, which is likely an error in the original question. Assuming D is meant to be the transition from \( n = 3 \) to \( n = 1 \), D represents the largest energy difference, followed by C (which is the same as D, again suggesting an error), then B, and finally A.
Conclusion:
The transition corresponding to the emission of the shortest wavelength is D (assuming it is intended to represent the transition from \( n = 3 \) to \( n = 1 \)) (Option 2).
