An electron in an excited state of L i ^ 2 + ion has angular momentum (3h)/(2π) . The de Broglie wavelength of the electron in this state is pπ α _ 0 (where a _ 0 is the Bohr radius). The value of p is

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From Bohr's law mvr = (nh)/(2π) = (3h)/(2π) (from eqes.) ⇒ n = 3 And momentum = mv = (3h)/(2 ) Now, radius of n ^ th shell, r = ( n ^ 2 z ) a _ 0 ⇒ r = (3) ^ 2 3 . a _ 0 [∵ Z _ Li = 3] ⇒ r = 3 a _ 0 From De Broglie law wavelength = (h)/(Momentum) ⇒ λ = (h)/(mv) = (h)/((3h)/(2π)r) ⇒ λ = (2 )/(3) = (2π)/(3) × 3 a _ 0 λ = 2π a _ 0 = pπ a _ 0 ⇒ P = 2

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