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According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?
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The correct answer is:
$n=6$ to $n=5$
$\because \Delta E \propto\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$, where $n_2>n_1$ Energy of photon obtained from the transition $n=6$ to $n=5$ will have least energy.
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