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Q.

Shortest wavelength in the Brackett series of $B e^{3 +}$ Ion is

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a

$\frac{1}{R_{H}}$

b

$\frac{1}{9 R_{H}}$

c

$\frac{25}{9 R_{H}}$

d

$\frac{4}{R_{H}}$

answer is A.

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Detailed Solution

$n_{1} = 4$ ; $n_{2} = \infty$

$\bar{υ} = R h Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R h {(4)}^{2} (\frac{1}{4^{2}} - \frac{1}{\infty^{2}})$

$\bar{υ} = R h (16) (\frac{1}{16}) = R h$

$λ = \frac{1}{\bar{υ}} = \frac{1}{R h}$

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