The Total solution for NCERT class 6-12
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Answer - 15 : - The maximum no. of emission lines = n(n–1)2 = 6(6–1)2 =3× 5 = 15
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(ii) For hydrogenatom ; rn = 0.529 x n2 År5 = 0.529 x (5)2 = 13.225 Å = 1.3225 nm.
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