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Q.

An electron is orbiting in 4th energy state in a hydrogen atom. The de-Broglie wavelength associated with this electron is nearly equal to

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a

1.32 Å

b

13.2 Å

c

15.6 Å

d

20 Å

answer is B.

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Detailed Solution

p=2mKE=2×9.1×10−31×0.85×1.6×10−19≈5×10−25 λ=hp=6.63×10−345×10−25≈13.2Å
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