First slide

Bohr Model of the Hydrogen atom.

Question

In Hydrogen spectrum the shortest wavelength in Balmer series is $λ$ . The shortest wavelength in the Brackett series will be

Moderate

A

$2 λ$
B

$4 λ$
C

$9 λ$
D

$16 λ$

Solution

For shortest wavelength in Balmer series, $n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R (\frac{1}{4^{2}} - \frac{1}{\infty}) \Rightarrow λ = \frac{4}{R}$
For shortest wavelength in Brackett series, $n_{1} = 4, n_{2} = \infty$
$∴ \frac{1}{λ'} = R (\frac{1}{4^{2}} - \frac{1}{\infty}) \Rightarrow λ' = \frac{16}{R} = 4 λ$

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