A) 3
B) 4
C) 5
D) 6
Correct Answer: C
Solution :
[c] \[E=-13.6\,{{Z}^{2}}/{{n}^{2}}=-\frac{13.6\times 4}{{{n}^{2}}}=\frac{54.4}{{{n}^{2}}}eV\] In coming to ground state \[\Delta E={{E}_{0}}-{{E}_{i}}\] \[=\left( \frac{-54.4}{{{n}^{2}}} \right)-\left( \frac{-54.4}{1} \right)\] \[=54.4\left( 1-\frac{1}{{{n}^{2}}} \right)eV\] ?. (1) But \[\Delta E={{E}_{ph}}=\frac{hc}{{{\lambda }_{1}}}+\frac{hc}{{{\lambda }_{2}}}\] \[=12400\left( \frac{1}{1085}+\frac{1}{304} \right)eV\] ?. (2) From (1) & (2) \[=54.4\left( 1-\frac{1}{{{n}^{2}}} \right)=52.08\] \[\Rightarrow 1-\frac{-1}{{{n}^{2}}}=0.96\Rightarrow n=5\]You need to login to perform this action.
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