Atomic StructureEasy
Question
The K.E. of electron in He+ will be maximum in
Options
A.third orbit
B.first orbit
C.seventh orbit
D.infinite orbit
Solution
$K.E. = \frac{1}{2}mv^{2} = \frac{1}{2}m\left( 2.188 \times 10^{6} \times \frac{z}{n} \right)^{2} \propto \frac{1}{n^{2}}$
∴ For maximum K.E., n = 1 (minimum)
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