First slide

Bohr Model of the Hydrogen atom.

Question

If the wavelengths of the first line of the Lyman series for the hydrogen atom is 1210 $Å$ then the wavelength of the first line of the Balmer series of the hydrogen spectrum is

Moderate

A

$6534 Å$
B

$1210 Å$
C

$2420 Å$
D

$3660 Å$

Solution

we know that
$\frac{1}{λ} = R [\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}] For first line of Lyman series, n_{1} = 1 and n_{2} = 2 ∣ \frac{1}{λ_{1}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} or R = \frac{4}{3 λ_{1}} For first lien of Balmer series n_{1} = 2 and n_{2} = 3 \frac{1}{λ_{2}} = R [\frac{1}{4} - \frac{1}{9}] = \frac{5 R}{36} = \frac{5}{36} [\frac{4}{3 λ_{1}}] or \frac{1}{λ_{2}} = \frac{5}{36} \times \frac{4}{3} \times \frac{1}{1210} or λ_{2} = \frac{36 \times 3 \times 1210}{5 \times 4} = 6534 Å$

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