A doubly ionised Lithium atom is excited from its ground state (n=1) to n=3 state. The wavelengths of the spectral lines for corresponding transition are given by λ32,λ31 and λ21. The value of λ32λ31 and Λ21λ31 are respectively

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8.1, 0.67

8.1, 1.2

6.4, 1.2

6.4, 0.67

answer is C.

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Detailed Solution

1λ=R1n12−1n22⇒1λ32=536R⇒λ32=365R Similarly λ31=98R&λ21=43R ∴ λ32λ31=6.4 and λ21λ31=1.2

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