Atomic StructureHard
Question
What is the distance travelled by an electron revolving in the second orbit of Be3+ ion in 100 revolutions?
Options
A.3.32 × 10–8 m
B.5.29 × 10–8 m
C.6.64 × 10–8 m
D.1.16 × 10–8 m
Solution
$d = 2\pi r \times 100 = 2\pi \times \left( 0.529 \times \frac{2^{2}}{4} \times 10^{- 10}\text{ m} \right) \times 100 = 3.32 \times 10^{- 8}\text{ m}$
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