In Bohr's atom, energy is absorbed or radiated when an electron makes a transition between two stationary states. If $f$ is the frequency of emitted radiation and $E_{1}$ and $E_{2}$ are the energies of electron in two states, then practically which is correct for a hydrogen like atom for the transition as shown in the figure.

h f = | E_{1} - E_{2} |

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h f < | E_{1} - E_{2} |

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h f > | E_{1} - E_{2} |

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| E_{1} - E_{2} | \leq h f \leq | E_{1} + E_{2} |

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Solution

The correct option is B $h f < | E_{1} - E_{2} |$
When an atom that is in excited state $E^{'}$ emits a photon of energy $h f$ , the photon also carries a momentum $P = \frac{h f}{c} = \frac{h}{λ}$ . Conservation of momentum requires that atom must recoil with a momentum $p$ and so, will have a kinetic energy $\frac{P^{2}}{2 m}$ where, $m =$ mass of atom.
So, conservation of energy gives.
$Δ E = E^{'} - E$
$= h f +$ recoil KE of atom
$\Rightarrow h f = Δ E - \frac{(P)^{2}}{2 m}$
$\Rightarrow h f < Δ E$

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