If the short series limit of the Balmer series for hydrogen is 3646 Å, calculate the atomic number of the element which gives X-ray wavelengths down to 1.0 Å. Identify the element.

Solution

Given, wavelength of the balmer series = 3646 Å

The short limit of the Balmer series is given by the Rydberg's formula,

\bar{v} = \frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) \bar{v} = \frac{R}{4}

∴

R = \frac{4}{λ} = \frac{4}{3646} \times 10^{10} m^{- 1}

Further the wavelengths of the K_α series are given by the relation,

\bar{v} = \frac{1}{λ} = R (Z - 1)^{2} (\frac{1}{1^{2}} - \frac{1}{n^{2}})

The maximum wave number corresponds to n = ∞ and, therefore, we must have

\bar{v} = \frac{1}{λ} = R (Z - 1)^{2}

\Rightarrow

(Z - 1)^{2} = \frac{1}{Rλ} = \frac{3646 \times 10^{- 10}}{4 \times 1 \times 10^{- 10}} = 911.5

∴

(Z - 1) = \sqrt{911.5}

\Rightarrow

≅ 30.2

\Rightarrow

Z = 31.2 ≅ 31

Thus, the atomic number of the element concerned is 31.

We can infer that the element having atomic number Z = 31 is Gallium.

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