Question

When the target material in x-ray setup is copper, then $K_{\alpha}$ line is having wavelength $\lambda_0$. Another $K_{\alpha}$ line is also observed with wavelength of $\frac{784}{625}\lambda_0$, then atomic number of impurity because of which another $K_{\alpha}$ line is observed is

• A. 24
• B. 26
• C. 28
• D. 30

Answer 1

Answer: (B)

26

Answer 2

Moseley's Law states that the frequency ($\nu$) of characteristic X-rays is related to the atomic number ($Z$) of the element by $\sqrt{\nu} = a(Z-b)$, where $a$ and $b$ are constants. For $K_\alpha$ lines, the screening constant $b \approx 1$. Since $\nu = c/\lambda$ (where $c$ is the speed of light and $\lambda$ is the wavelength), the law can be written as $\frac{1}{\sqrt{\lambda}} = K(Z-1)$, or $\frac{1}{\lambda} = K^2(Z-1)^2$, where $K^2$ is a new constant.

For copper ($Z_{Cu} = 29$) with $K_\alpha$ wavelength $\lambda_0$: $\frac{1}{\lambda_0} = K^2(Z_{Cu}-1)^2 = K^2(29-1)^2 = K^2(28)^2$

For the impurity with atomic number $Z_{imp}$ and $K_\alpha$ wavelength $\lambda_{imp} = \frac{784}{625}\lambda_0$: $\frac{1}{\lambda_{imp}} = K^2(Z_{imp}-1)^2$

Dividing the equation for the impurity by the equation for copper: $\frac{1/\lambda_{imp}}{1/\lambda_0} = \frac{K^2(Z_{imp}-1)^2}{K^2(Z_{Cu}-1)^2}$ $\frac{\lambda_0}{\lambda_{imp}} = \frac{(Z_{imp}-1)^2}{(Z_{Cu}-1)^2}$

Substitute the given values: $\lambda_{imp} = \frac{784}{625}\lambda_0$, so $\frac{\lambda_0}{\lambda_{imp}} = \frac{625}{784}$, and $Z_{Cu} = 29$. $\frac{625}{784} = \frac{(Z_{imp}-1)^2}{(29-1)^2}$ $\frac{625}{784} = \frac{(Z_{imp}-1)^2}{(28)^2}$

Take the square root of both sides: $\sqrt{\frac{625}{784}} = \frac{Z_{imp}-1}{28}$ $\frac{25}{28} = \frac{Z_{imp}-1}{28}$

Solving for $Z_{imp}$: $25 = Z_{imp}-1$ $Z_{imp} = 25 + 1 = 26$

The atomic number of the impurity is 26.