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14/08/2021 11:32 am
Topic starter
What is the angular velocity (w) of an electron occupying second orbit of Li2+ ion?
(a) \(\frac{8 \pi^3 me^4}{h^3}K^2\)
(b) \(\frac{8 \pi^3 me^4}{9h^3}K^2\)
(c) \(\frac{64}{9} \times \frac{ \pi^3 me^4}{h^3}K^2\)
(d) \(\frac{9 \pi^3 me^4}{h^3}K^2\)
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14/08/2021 11:43 am
Topic starter
Correct answer:(d) \(\frac{9 \pi^3 me^4}{h^3}K^2\)
Explanation:
vn = rn ω where rn= \(\frac{n^2h^2}{4 \pi^2 me^2Z.K}\)
and rn = \(\frac{2 \pi. Z.e^2.K}{n.h}\)
∴ \(\frac{2 \pi. Z.e^2.K}{n.h}\) = \(\frac{4 \pi^3 me^4}{h^3}K^2\) x ω
ω = \(\frac{8 \pi^3 me^4.Z^2.K^2}{n^3.h^3}K^2\)
= \(\frac{9 \pi^3 me^4}{h^3}K^2\)(∵ n = 2 and Z = 3)