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An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of the electron in this state is Pπa0 (where a0 is the Bohr radius). The value of P is

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answer is 2.

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Detailed Solution

Bohr's Postulate ,nh2π=3h2π⇒n=3Radius=r=a0n2z⇒r=a093=3a0Wavelength=λ=hmv=hrmvr=hrnh2π=2πrn⇒λ=2π3a03 ⇒λ=2πa0So, P=2

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