Frequency of a photon emitted due to transition of electron of a certain element from `L to K `shell is found to be `4.2 xx 10^(18) Hz` Using Moseley 's law, find the atomic number of the element , given that the Rydberg's constant `R = 1.1 xx 10^(7) m^(-1)`

Answer

To solve the problem, we will use Moseley’s law, which relates the frequency of emitted photons during electronic transitions to the atomic number of the element. The formula according to Moseley’s law is given by:

\[ \frac{1}{\lambda} = R \left( Z - 1 \right)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted photon
- \( R \) is the Rydberg constant
- \( Z \) is the atomic number of the element
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states of the electron transition.

In this case, the transition is from the L shell (n=2) to the K shell (n=1), so we can substitute \( n_1 = 1 \) and \( n_2 = 2 \).

### Step 1: Calculate the wavelength from frequency
We know that the frequency \( f \) is related to the wavelength \( \lambda \) by the equation:

\[ c = \lambda f \]

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

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

Substituting the given values:

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

Calculating this gives:

\[ \lambda \approx 7.14 \times 10^{-11} \, \text{m} \]

### Step 2: Substitute into Moseley’s Law
Now, we can substitute \( \lambda \) back into Moseley’s law:

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

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

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

Thus, we have:

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

### Step 3: Calculate \( \frac{1}{\lambda} \)
Now we calculate \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{1}{7.14 \times 10^{-11}} \approx 1.4 \times 10^{10} \, \text{m}^{-1} \]

### Step 4: Substitute Rydberg Constant
Substituting \( R = 1.1 \times 10^7 \, \text{m}^{-1} \):

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

### Step 5: Solve for \( Z \)
Rearranging gives:

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

Calculating the right side:

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

Taking the square root:

\[ Z - 1 \approx \sqrt{1.696 \times 10^3} \approx 41.2 \]

Thus:

\[ Z \approx 42.2 \approx 42 \]

### Conclusion
The atomic number \( Z \) of the element is approximately \( 42 \).

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