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Question

An electron in an excited state of Li2+ ion has angular momentum 3h2π. The de Broglie wavelength of the electron in this state in pπα0 (where α0 is the Bohr radius). The value of p is

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Solution

Angular momentum, L=nh2π=3h2π

n=3

Also, Wavelength λ=hp=hrmvr=hr3h2π=2πr3 ...(1)

But, r=a0n2Z ....(2)

Substituting (2) in (1), λ=2π3a0n2Z=2π3a0323=2πa0

p=2 comparing with the given expression pπa0

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