What is the approximate ratio of wavelength of radiation `K_(alpha_(1)) and K_(alpha_(2))` when atomic number of first target is `Z_(1) = 64` and atomic no of second target is `Z_(2)=80` -

Options

A. `(1)/(4)`
B. `(1)/(16)`
C. `(2)/(sqrt(5))`
D. `(25)/(16)`

Answer

To find the approximate ratio of the wavelengths of radiation \( K_{\alpha_1} \) and \( K_{\alpha_2} \) for atomic numbers \( Z_1 = 64 \) and \( Z_2 = 80 \), we can follow these steps:

### Step 1: Understand the relationship between wavelength and atomic number
According to Moseley's law, the frequency \( \nu \) of the emitted radiation is given by:
\[ \nu = k(Z - 1)^2 \]
where \( k \) is a constant and \( Z \) is the atomic number. The wavelength \( \lambda \) is related to frequency by the equation:
\[ \lambda = \frac{c}{\nu} \]
where \( c \) is the speed of light.

### Step 2: Express wavelength in terms of atomic number
From the relationship above, we can express the wavelength in terms of the atomic number:
\[ \lambda \propto \frac{1}{(Z - 1)^2} \]
This means that the wavelength is inversely proportional to the square of \( (Z - 1) \).

### Step 3: Set up the ratio of wavelengths
We want to find the ratio of the wavelengths \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{(Z_2 - 1)^2}{(Z_1 - 1)^2} \]

### Step 4: Substitute the values of \( Z_1 \) and \( Z_2 \)
Substituting \( Z_1 = 64 \) and \( Z_2 = 80 \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{(80 - 1)^2}{(64 - 1)^2} = \frac{79^2}{63^2} \]

### Step 5: Calculate the squares
Calculating the squares:
\[ 79^2 = 6241 \]
\[ 63^2 = 3969 \]

### Step 6: Calculate the ratio
Now, we can compute the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{6241}{3969} \approx 1.57 \]

### Conclusion
Thus, the approximate ratio of the wavelengths \( \lambda_1 \) and \( \lambda_2 \) is:
\[ \frac{\lambda_1}{\lambda_2} \approx 1.57 \]