An electron in an excited state of L i 2 + ion has angular momentum 3 h 2 π . The de Broglie wavelength of the electron in this state is [ a 0 is the Bohr radius]

Step-by-step solution

From Bohr's law m v r = n h 2 π = 3 h 2 π ⇒ n = 3 And momentum = m v = 3 h 2 π r Now, radius of n t h shell, r = n 2 z a 0 ⇒ r = 3 2 3 . a 0 ∵ Z L i = 3 ⇒ r = 3 a 0 From De Broglie law w a v e l e n g t h = h M o m e n t u m ⇒ λ = h m v = h 3 h 2 π r ⇒ λ = 2 π r 3 = 2 π 3 × 3 a 0 λ = 2 π a 0