Question

Characteristic X-rays of frequency 4.2 x10$^{18}$ Hz are produced when transitions from $L$ shell take place in a certain target material. Use Moseley's law and determine the atomic number of the target material. Given, Rydberg constant is R = 1.1 x 10$^{7}$/m

Answer 1

42

Answer 2

To determine the atomic number of the target material, we use Moseley's Law for characteristic X-rays.

Moseley's Law states that the frequency ($\nu$) of a characteristic X-ray line is related to the atomic number ($Z$) by the formula:

$\nu = R c (Z-b)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where:

• $R$ is the Rydberg constant.
• $c$ is the speed of light.
• $Z$ is the atomic number.
• $b$ is the screening constant.
• $n_1$ and $n_2$ are the principal quantum numbers of the inner and outer shells involved in the electron transition, respectively.

The problem states that "transitions from L shell take place". This typically refers to an electron from the L-shell (n=2) filling a vacancy in the K-shell (n=1). This transition produces a K-alpha ($K_{\alpha}$) X-ray. For K-alpha lines, the screening constant $b$ is approximately 1.

So, for a K-alpha transition:

$n_1 = 1$ (K-shell)

$n_2 = 2$ (L-shell)

$b = 1$

Substituting these values into Moseley's Law:

$\nu = R c (Z-1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$

$\nu = R c (Z-1)^2 \left( 1 - \frac{1}{4} \right)$

$\nu = R c (Z-1)^2 \left( \frac{3}{4} \right)$

Given values:

Frequency $\nu = 4.2 \times 10^{18}$ Hz

Rydberg constant $R = 1.1 \times 10^{7}$ /m

Speed of light $c = 3 \times 10^8$ m/s

Now, we can plug in the values and solve for $Z$:

$4.2 \times 10^{18} = (1.1 \times 10^{7} \text{ m}^{-1}) \times (3 \times 10^8 \text{ m/s}) \times (Z-1)^2 \times \frac{3}{4}$

First, calculate the product $R \times c$:

$R c = (1.1 \times 10^{7}) \times (3 \times 10^8) = 3.3 \times 10^{15}$ Hz

Substitute this back into the equation:

$4.2 \times 10^{18} = (3.3 \times 10^{15}) \times (Z-1)^2 \times \frac{3}{4}$

Rearrange the equation to solve for $(Z-1)^2$:

$(Z-1)^2 = \frac{4.2 \times 10^{18}}{3.3 \times 10^{15} \times \frac{3}{4}}$

$(Z-1)^2 = \frac{4.2 \times 10^{18}}{ (3.3 \times 10^{15}) \times 0.75 }$

$(Z-1)^2 = \frac{4.2 \times 10^{18}}{2.475 \times 10^{15}}$

$(Z-1)^2 = \frac{4.2}{2.475} \times 10^{(18-15)}$

$(Z-1)^2 = 1.696969... \times 10^3$

$(Z-1)^2 = 1696.969...$

Now, take the square root of both sides:

$Z-1 = \sqrt{1696.969...}$

$Z-1 \approx 41.194$

Finally, solve for $Z$:

$Z \approx 41.194 + 1$

$Z \approx 42.194$

Since the atomic number must be an integer, we round to the nearest whole number.

$Z = 42$

The target material is Molybdenum (Mo), which has an atomic number of 42.