Which of the following transitions will emit | vedclass

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(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = h 
u = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Del

Vedclass · Accepted Answer

$n_4 	o n_1$

Vedclass · Answer

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = h 
u = \frac{hc}{\lambda}$.Since $\lambda = \frac{hc}{\Delta E}$,the shortest wavelength corresponds to the transition with the largest energy difference $(\Delta E)$.Comparing the transitions:$(A)$ $n_4 	o n_1$: Largest energy gap.$(B)$ $n_2 	o n_1$: Smaller energy gap.$(C)$ $n_4 	o n_2$: Smaller energy gap.$(D)$ $n_3 	o n_1$: Smaller energy gap than $n_4 	o n_1$.Therefore,the transition $n_4 	o n_1$ involves the maximum energy change,resulting in the shortest wavelength.

Which of the following transitions will emit the shortest wavelength?

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