A) 13.2 E
B) 7.2 E
C) 5.6 E
D) 3.2 E
Correct Answer: B
Solution :
Energy \[=0.0258\ amu\] (K = constant) n1 = 2 and n2 = 3, so \[E=K\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]=K\left[ \frac{5}{36} \right]\] For removing an electron n1 = 1 to \[{{n}_{2}}=\infty \] Energy \[{{E}_{1}}=K[1]=\frac{36}{5}E=7.2\ E\] \[\therefore \]Ionization energy = 7.2 EYou need to login to perform this action.
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