The equation corresponding to the wave number of spectral line in the Brackett series:

R [(1 / 2^{2}) - (1 / 4^{2})]

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R [(1 / 4^{2}) - (1 / 5^{2})]

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R [(1 / 3^{2}) - (1 / 5^{2})]

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R [(1 / 5^{2}) - (1 / 6^{2})]

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Solution

The correct option is A $R [(1 / 4^{2}) - (1 / 5^{2})]$
Brackett series is for electronic transition from n $= 4$ to some higher energy level .
Also , the wavenumber for any electronic transition is never negative.
Use the formula,
$1 / λ = R \times (1 / n_{1}^{2} - 1 / n_{2}^{2})$

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\frac{1}{3^{2} + 1} + \frac{1}{4^{2} + 2} + \frac{1}{5^{2} + 3} + \frac{1}{6^{2} + 4} + . . . \infty

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\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . . . \infty

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\frac{1}{3^{2} + 1} + \frac{1}{4^{2} + 2} + \frac{1}{5^{2} + 3} + \frac{1}{6^{2} + 4}

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x = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . .,

y = \frac{1}{1^{2}} + \frac{3}{2^{2}} + \frac{1}{3^{2}} + \frac{3}{4^{2}} + . .

z = \frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . .,

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