Use Moseley's law with b = 1 to find the frequency of the `K_alpha` X-ray of `La(Z = 57)` if the frequency of the `K_alpha` X-ray of `Cu(Z = 29)` is known to be `1*88 xx 10^(18)` Hz.

Explanation

To find the frequency of the `K_alpha` X-ray of Lanthanum (La, Z = 57) using Moseley's law, we will follow these steps:

### Step 1: Write down Moseley's law
Moseley's law states that the frequency of the X-ray emitted is proportional to the square of the atomic number (Z) minus a constant (b):
\[ \nu \propto (Z - b)^2 \]
In this case, we are given \( b = 1 \).

### Step 2: Express the frequency for Lanthanum
For Lanthanum (Z = 57), we can express the frequency as:
\[ \nu_{La} = a \cdot (Z_{La} - b) = a \cdot (57 - 1) = a \cdot 56 \]
This can be rewritten as:
\[ \nu_{La} = a \cdot 56 \tag{1} \]

### Step 3: Express the frequency for Copper
For Copper (Cu, Z = 29), we know the frequency is given as \( \nu_{Cu} = 1.88 \times 10^{18} \) Hz. Using Moseley's law, we can express the frequency for Copper as:
\[ \nu_{Cu} = a \cdot (Z_{Cu} - b) = a \cdot (29 - 1) = a \cdot 28 \]
This can be rewritten as:
\[ \nu_{Cu} = a \cdot 28 \tag{2} \]

### Step 4: Relate the two equations
Now, we can relate the two equations (1) and (2) by dividing them:
\[ \frac{\sqrt{\nu_{Cu}}}{\sqrt{\nu_{La}}} = \frac{a \cdot 28}{a \cdot 56} \]
The \( a \) cancels out:
\[ \frac{\sqrt{\nu_{Cu}}}{\sqrt{\nu_{La}}} = \frac{28}{56} = \frac{1}{2} \]

### Step 5: Solve for \( \nu_{La} \)
Now we can square both sides to eliminate the square root:
\[ \frac{\nu_{Cu}}{\nu_{La}} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Thus, we can express \( \nu_{La} \) in terms of \( \nu_{Cu} \):
\[ \nu_{La} = 4 \cdot \nu_{Cu} \]
Substituting the value of \( \nu_{Cu} \):
\[ \nu_{La} = 4 \cdot (1.88 \times 10^{18}) = 7.52 \times 10^{18} \text{ Hz} \]

### Final Answer
The frequency of the `K_alpha` X-ray of Lanthanum is:
\[ \nu_{La} = 7.52 \times 10^{18} \text{ Hz} \]