A piece of tissue paper soaked in polluted water is irradiated with continuous X-rays. The incident X-rays knock electrons out of the inner shells of the sample's atoms, causing excitation. Subsequently, characteristic

Question

Quantrex Academy · Accepted Answer

Correct answer: D. Zr, Zn, Cu, Fe

Using Moseley's equation given in the problem:

\nu = (25 \times 10^{14} \text{ Hz})(Z - 1)^2

We know the relation between frequency

\nu

and wavelength

\lambda

is

\nu = \dfrac{c}{\lambda}

, where

c = 3 \times 10^8 \text{ m/s}

.

Substituting

\nu

into the equation and converting

\lambda

from pm to meters (

\lambda \times 10^{-12} \text{ m}

):

\dfrac{3 \times 10^8}{\lambda \times 10^{-12}} = 25 \times 10^{14} (Z - 1)^2

Rearranging to solve for

(Z - 1)^2

:

(Z - 1)^2 = \dfrac{3 \times 10^{20}}{25 \times 10^{14} \times \lambda} = \dfrac{120000}{\lambda}

Z = 1 + \sqrt{\dfrac{120000}{\lambda}}

Now, we calculate the atomic number

Z

for each peak wavelength from the given graph:

For

\lambda = 78.9 \text{ pm}

:

Z = 1 + \sqrt{\dfrac{120000}{78.9}} \approx 1 + 39.0 = 40 \Rightarrow \text{Zr (Zirconium)}

For

\lambda = 146 \text{ pm}

:

Z = 1 + \sqrt{\dfrac{120000}{146}} \approx 1 + 28.67 \approx 30 \Rightarrow \text{Zn (Zinc)}

For

\lambda = 158 \text{ pm}

:

Z = 1 + \sqrt{\dfrac{120000}{158}} \approx 1 + 27.56 \approx 29 \Rightarrow \text{Cu (Copper)}

For

\lambda = 198 \text{ pm}

:

Z = 1 + \sqrt{\dfrac{120000}{198}} \approx 1 + 24.62 \approx 26 \Rightarrow \text{Fe (Iron)}

The elements present in the sample are Zr, Zn, Cu, and Fe.

Answer: Zr, Zn, Cu, Fe

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