if `lambda_(Cu)` is the wavelength of `K_alpha`, X-ray line fo copper (atomic number 29) and `lambda_(Mo)` is the wavelength of the `K_alpha` X-ray line of molybdenum (atomic number 42), then the ratio `lambda_(Cu)/lambda_(Mo)`is close to
(a) 1.99 (b) 2.14
(c ) 0.50 (d) 0.48

Explanation

To solve the problem of finding the ratio \( \frac{\lambda_{Cu}}{\lambda_{Mo}} \) for the K-alpha X-ray lines of copper and molybdenum, we can follow these steps:

### Step 1: Understand the relationship between wavelength and atomic number
The wavelength of the K-alpha X-ray line is related to the atomic number of the element. According to Moseley's law, the frequency of the X-ray emitted is proportional to the square of the atomic number minus a constant. The relationship can be expressed as:
\[ \nu \propto (Z - \beta)^2 \]
where \( Z \) is the atomic number and \( \beta \) is a constant (approximately equal to 1 for K-alpha lines).

### Step 2: Relate wavelength to frequency
Since wavelength \( \lambda \) is inversely proportional to frequency \( \nu \), we can write:
\[ \lambda \propto \frac{1}{\nu} \]
This means:
\[ \lambda \propto \frac{1}{(Z - \beta)^2} \]

### Step 3: Set up the ratio of wavelengths
For copper (Cu) and molybdenum (Mo), we can express the ratio of their wavelengths as:
\[ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{(Z_{Mo} - \beta)^2}{(Z_{Cu} - \beta)^2} \]

### Step 4: Substitute the atomic numbers
Substituting the atomic numbers:
- For copper, \( Z_{Cu} = 29 \)
- For molybdenum, \( Z_{Mo} = 42 \)

Using \( \beta \approx 1 \):
\[ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{(42 - 1)^2}{(29 - 1)^2} = \frac{41^2}{28^2} \]

### Step 5: Calculate the squares
Calculating the squares:
\[ 41^2 = 1681 \]
\[ 28^2 = 784 \]

### Step 6: Calculate the ratio
Now, calculate the ratio:
\[ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{1681}{784} \]

### Step 7: Simplify the ratio
To simplify:
\[ \frac{1681}{784} \approx 2.14 \]

### Conclusion
Thus, the ratio \( \frac{\lambda_{Cu}}{\lambda_{Mo}} \) is approximately \( 2.14 \). Therefore, the correct answer is (b) 2.14.

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