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A single hydrogen atom in ground state is given least energy $E_{0}$ so that it can emit $H_{β}$ - line in the Balmer series, then :

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$E_{0} \approx 12.09 e V$

$E_{0} \approx 12.75 e V$

This atom can give maximum 3 different lines in emission spectrum.

This atom can give maximum 6 different lines in emission spectrum.

answer is B, C.

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Detailed Solution

Transition that should occur is $n = 1$ to $n = 4$
$\Rightarrow E_{0} = E_{4} - E_{1} = (- 0.85 + 13.6) e V = 12.75 e V$
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Maximum 3 different lines can be emitted in emission spectrum as shown

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