Atomic StructureEasy
Question
As the orbit number increases, the K.E. and P.E. for an electron
Options
A.both increases.
B.both decreases.
C.K.E. increases but P.E. decreases.
D.P.E. increases but K.E. decreases.
Solution
$K.E. = \frac{1}{2}mv^{2},P.E. = - mv^{2}\text{ and }V \propto \frac{1}{n}$
Therefore, with increase in orbit number, K.E. decreased but P.E. increases.
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