A) \[{{n}_{1}}=2\to {{n}_{2}}=1\]
B) \[{{n}_{1}}=3\to {{n}_{2}}=1\]
C) \[{{n}_{1}}=10\to {{n}_{2}}=2\]
D) \[{{n}_{1}}=100\to {{n}_{2}}=3\]
Correct Answer: B
Solution :
\[{{E}_{1}}=-13.6\text{ }eV,\text{ }{{E}_{3}}=-\text{ }3.4\text{ }eV,\text{ }{{E}_{3}}=-1.51\text{ }eV,\] \[{{E}_{10}}=-\text{ }0.136\text{ }eV,\text{ }{{\text{E}}_{100}}=-0.0013\text{ }eV\] \[{{E}_{21}}=-13.6-(-3.4)=-10.2\text{ }eV\] \[{{E}_{31}}=-13.6-(-1.51)=-12.09\text{ }eV\] \[{{E}_{10,2}}=-3.4-(-0.136)=-\text{ }3.264\text{ }eV\] \[{{E}_{100,3}}=-1-51-(-0.0013)=-1.508eV\] Hence,\[{{E}_{31}}\]is the most energetic transition.
You need to login to perform this action.
You will be redirected in
3 sec
