A) may emit another photon in the Paschen series
B) need not emit any more photon
C) may emit another photon in the Balmer series
D) must emit another photon in the Lyman series.
Correct Answer: D
Solution :
\[\frac{1}{\lambda }=R\left( \frac{1}{n{{f}^{2}}}-\frac{1}{n_{t}^{2}} \right)\]Balmer series lies in emission spectra. \[{{n}_{i}}=3\] \[{{n}_{f}}=2\] When electron jumps from n = 3 to n = 2 and emit a photon. Then to fill of this vacancy created by an \[\overline{e}\] jump another i from any excited state jump to n = 1 state and there by emitted a photon whose energy is equal to the difference between the two levels. After \[{{10}^{-8}}\] sec both return to their original position. Then also emit photons process continue and we observe line spectrum. Hence one photon emitted in Balmer series. Then \[{{H}_{2}}\] atom must emit another photon in Lymen series.
You need to login to perform this action.
You will be redirected in
3 sec
