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Q.

The debroglie wave length of an electron in an orbit represented by the orbital diagram

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a

6.66A°

b

10A°

c

20A°

d

13.33A°

answer is A.

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Detailed Solution

No. of electron waves=n=3

$2 πr = nλ λ = \frac{2 πr}{n} = \frac{2 \times 3.14 \times 0.53 \times 3^{2}}{3} \approx 10 A^{o}$

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