Atomic StructureHard

Question

The ratio of time taken by electron in revolution around the H-nucleus in the second and third orbits is

Options

A.2: 3
B.4: 8
C.8: 27
D.27: 8

Solution

$T_{n,z} = 1.5 \times 10^{- 16} \times \frac{n^{3}}{z^{2}}\text{ sec}$

$\therefore\frac{T_{2,M}}{T_{3,M}} = \frac{2^{3}}{3^{3}} = \frac{8}{27}$

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