Characteristic X-rays of frequency `4.2xx10^18` Hz are produced when transitions from L-shell to K-shell take place in a certain target material. Use Mosley's law to determine the atomic number of the target material. Given Rydberg constant `R =1.1xx10^7 m^(-1)`.

Explanation

To determine the atomic number of the target material using Moseley's law, we can follow these steps:

### Step 1: Understand Moseley's Law
Moseley's law relates the wavelength (or frequency) of the emitted X-rays to the atomic number (Z) of the element. The formula is given by:

\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \cdot \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Where:
- \(\lambda\) is the wavelength of the emitted X-rays,
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_f\) is the principal quantum number of the final state,
- \(n_i\) is the principal quantum number of the initial state.

### Step 2: Convert Frequency to Wavelength
We know the frequency \(f\) of the X-rays is given as \(4.2 \times 10^{18}\) Hz. We can use the relationship between frequency and wavelength:

\[ \lambda = \frac{c}{f} \]

Where \(c\) is the speed of light (\(c \approx 3 \times 10^8 \, \text{m/s}\)).

Calculating the wavelength:

\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{4.2 \times 10^{18} \, \text{Hz}} \approx 7.14 \times 10^{-11} \, \text{m} \]

### Step 3: Substitute Values into Moseley's Law
In this case, the transition is from the L-shell (n_i = 2) to the K-shell (n_f = 1). Thus, we have:

\[ n_f = 1, \quad n_i = 2 \]

Substituting these values into the formula:

\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \cdot \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating \(\left( \frac{1}{1^2} - \frac{1}{2^2} \right)\):

\[ \frac{1}{1} - \frac{1}{4} = 1 - 0.25 = 0.75 \]

### Step 4: Substitute Rydberg Constant and Wavelength
Now, substituting the values into the equation:

\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \cdot 0.75 \]

Using \(R = 1.1 \times 10^7 \, \text{m}^{-1}\) and \(\lambda \approx 7.14 \times 10^{-11} \, \text{m}\):

\[ \frac{1}{7.14 \times 10^{-11}} = 1.1 \times 10^7 \cdot (Z - 1)^2 \cdot 0.75 \]

Calculating \(\frac{1}{7.14 \times 10^{-11}}\):

\[ \approx 1.4 \times 10^{10} \, \text{m}^{-1} \]

### Step 5: Solve for \(Z\)
Now we can set up the equation:

\[ 1.4 \times 10^{10} = 1.1 \times 10^7 \cdot (Z - 1)^2 \cdot 0.75 \]

Dividing both sides by \(1.1 \times 10^7 \cdot 0.75\):

\[ (Z - 1)^2 = \frac{1.4 \times 10^{10}}{1.1 \times 10^7 \cdot 0.75} \]

Calculating the right-hand side:

\[ (Z - 1)^2 \approx \frac{1.4 \times 10^{10}}{0.825 \times 10^7} \approx 1.696 \times 10^3 \approx 1696 \]

Taking the square root:

\[ Z - 1 \approx 41.2 \]

Thus,

\[ Z \approx 42.2 \]

Since \(Z\) must be an integer, we round it to:

\[ Z = 42 \]

### Final Answer
The atomic number of the target material is \(Z = 42\).

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