Atomic StructureHard

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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