Chapter 28, Problem 36P

To determine

The unknown material.

Answer to Problem 36P

Chromium

Explanation of Solution

Given info:

The wavelength from unknown material is ${\text{λ}}_{\text{X}}\text{=229}\text{pm}\text{=}{\text{229×10}}^{\text{-12}}\text{m}$

The wavelength of ${\text{K}}_{\alpha }$ line is ${\text{λ}}_{\text{iron}}\text{=}\text{194}\text{pm}\text{=194}{\text{×10}}^{\text{-12}}\text{m}$

Formula used:

  $\text{λ}\propto \frac{1}{{\left(Z-1\right)}^2}$

Where,

  $\text{λ}$ is a wavelength and $\text{Z}$ is atomic number.

Calculation:

The ${\text{K}}_{\alpha }$ line is from the $\text{n = 2}$ to $\text{n = 1}$ transition.

Use the energies of the hydrogen atom with $\text{Z}$ replaced by $\text{Z-1}$

  $hv\text{=}\text{ΔE}\propto {\left(\text{Z-1}\right)}^{\text{2}}$

So, $\text{λ}\propto \frac{1}{{\left(Z-1\right)}^2}$

Now, form the ratio for the two materials:

  $\frac{{\text{λ}}_{\text{X}}}{{\text{λ}}_{\text{iron}}}\text{=}\frac{{\left({\text{Z}}_{\text{iron}}\text{-1}\right)}^{\text{2}}}{{\left({\text{Z}}_{\text{x}}\text{-1}\right)}^{\text{2}}}$

  $\frac{\left(\text{229}\text{nm}\right)}{\left(\text{194}\text{nm}\right)}\text{=}\frac{{\left(\text{26-1}\right)}^{\text{2}}}{{\left({\text{Z}}_{\text{x}}\text{-1}\right)}^{\text{2}}}$

This gives ${\text{Z}}_{\text{x}}=24$

So, the material is chromium.

Conclusion:

The required material is chromium.