A) \[-6.8eV\]
B) \[-4.2eV\]
C) \[-3.4eV\]
D) \[+6.8eV\]
Correct Answer: C
Solution :
\[{{E}_{1}}\] for \[H=-13.6eV.\] \[{{E}_{n}}\] for \[H=\frac{{{E}_{1}}}{{{n}^{2}}e}\] Thus, \[{{E}_{2}}=\frac{-13.6}{4}=-3.4eV\] \[{{E}_{3}}=\frac{-13.6}{9}=-1.51eV\] \[{{E}_{4}}=\frac{-13.6}{16}=-0.85\] and so onYou need to login to perform this action.
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