Out of the following transitions in a hydroge | vedclass

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(B) The frequency of an emitted photon during an electronic transition is given by the relation $E = h
u = E_i - E_f$,where $E_i$ is the initial ener

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$n=2$ to $n=1$

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(B) The frequency of an emitted photon during an electronic transition is given by the relation $E = h
u = E_i - E_f$,where $E_i$ is the initial energy level and $E_f$ is the final energy level.For emission to occur,the electron must transition from a higher energy level to a lower energy level $(n_i > n_f)$.Thus,options $A$ ($n=1$ to $n=2$) and $C$ ($n=2$ to $n=6$) involve absorption,not emission,and can be eliminated.Comparing the emission transitions:$1$. For $n=2$ to $n=1$: $\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 	imes 0.75 = 10.2 	ext{ eV}$.$2$. For $n=6$ to $n=2$: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{6^2} ight) = 13.6 \left( 0.25 - 0.0277 ight) = 13.6 	imes 0.2223 \approx 3.02 	ext{ eV}$.Since frequency $
u = \frac{\Delta E}{h}$,the transition with the largest energy difference $\Delta E$ will emit the photon with the highest frequency.Therefore,the transition $n=2$ to $n=1$ emits the photon of the highest frequency.

Out of the following transitions in a hydrogen atom,identify the transition which emits photons of highest frequency.

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