The wavelength of `K_(alpha)` line for an elèment of atomic number 29 is `lambda`. The wavelength of `K_(alpha)` line for an element of atomic number 15 is `beta lambda .` Find `beta .` Take Moseley's constant, `b=1` for both elements)

Answer

To solve the problem, we will use the formula for the wavelength of the K-alpha line, which is given by:

\[ \lambda \propto \frac{1}{(Z - 1)^2} \]

where \( Z \) is the atomic number of the element.

### Step 1: Write the relationship for the element with atomic number 29
For the element with atomic number \( Z_1 = 29 \):
\[ \lambda_1 = \lambda \propto \frac{1}{(29 - 1)^2} = \frac{1}{28^2} \]

### Step 2: Write the relationship for the element with atomic number 15
For the element with atomic number \( Z_2 = 15 \):
\[ \lambda_2 = \beta \lambda \propto \frac{1}{(15 - 1)^2} = \frac{1}{14^2} \]

### Step 3: Set up the proportionality relationship
From the above relationships, we can express the proportionality as:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{1}{14^2}}{\frac{1}{28^2}} = \frac{28^2}{14^2} \]

### Step 4: Simplify the ratio
Now, simplifying the ratio:
\[ \frac{28^2}{14^2} = \left(\frac{28}{14}\right)^2 = 2^2 = 4 \]

### Step 5: Relate the wavelengths
Since we have:
\[ \frac{\lambda_2}{\lambda_1} = \beta \]
We can substitute:
\[ \beta = 4 \]

### Final Answer
Thus, the value of \( \beta \) is:
\[ \beta = 4 \]

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