The absorption transition between the ground state and excited state of hydrogen atom is 3. The emission transitions between these states will be

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Solution

The correct option is D 6
Number of absorption lines = (n – 1) $\Rightarrow$ 3 = (n – 1) $\Rightarrow$ n = 4
Hence number of emitted lines $= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$

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The absorption transition between the ground state and excited state of hydrogen atom is 3. The emission transitions between these states will be

The correct option is D 6Number of absorption lines = (n – 1) ⇒ 3 = (n – 1) ⇒ n = 4 Hence number of emitted lines =n(n−1)2=4(4−1)2=6

The correct option is D 6
Number of absorption lines = (n – 1) $\Rightarrow$ 3 = (n – 1) $\Rightarrow$ n = 4
Hence number of emitted lines $= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$