A) from n = 2 to n = 1
B) from n = 3 to n = 1
C) from n = 4 to n = 2
D) from n = 3 to n = 2
Correct Answer: B
Solution :
For hydrogen atom change in energy \[\Delta E=Rhc\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] Going through options, we get \[\Delta {{E}_{1}}=Rhc\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3}{4}Rhc=0.75Rhc\] \[\Delta {{E}_{2}}=Rhc\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{8}{9}Rhc=0.8\bar{8}Rhc\] \[\Delta {{E}_{3}}=Rhc\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3}{16}Rhc=0.187Rhc\] \[\Delta {{E}_{4}}=Rhc\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5}{36}Rhc=0.138Rhc\] So, energy change is maximum in option ie, \[\Delta {{E}_{2}}=0.8\bar{8}Rhc\]
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