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, we need to find the ratio of the wavelengths of the K_alpha X-ray lines for copper and molybdenum using the Rydberg formula for X-ray emissions. Here’s a step-by-step solution:

### Step 1: Understand the Rydberg Formula for X-rays
The wavelength of X-ray lines can be described by the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron transition.

### Step 2: Identify the Values for K_alpha Transition
For the K_alpha transition:
- \( n_1 = 1 \) (the lower energy level),
- \( n_2 = 2 \) (the higher energy level).

### Step 3: Write the Ratio of Wavelengths
We need to find the ratio \( \frac{\lambda_{Cu}}{\lambda_{Mo}} \). According to the Rydberg formula, we can express this ratio as:
\[ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{(Z_{Mo} - 1)^2}{(Z_{Cu} - 1)^2} \]

### Step 4: Substitute the Atomic Numbers
Given:
- Atomic number of copper \( Z_{Cu} = 29 \),
- Atomic number of molybdenum \( Z_{Mo} = 42 \).

Substituting these values into the equation:
\[ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \frac{(42 - 1)^2}{(29 - 1)^2} = \frac{41^2}{28^2} \]

### Step 5: Calculate the Squares
Calculating the squares:
- \( 41^2 = 1681 \),
- \( 28^2 = 784 \).

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

### Step 7: Simplify the Ratio
Calculating the numerical value:
\[ \frac{1681}{784} \approx 2.144 \]

### Step 8: Identify the Closest Option
From the options provided:
- (a) 1.99
- (b) 2.14
- (c) 0.50
- (d) 0.48

The closest option to our calculated value of approximately \( 2.144 \) is (b) 2.14.

### Final Answer
Thus, the ratio \( \frac{\lambda_{Cu}}{\lambda_{Mo}} \) is close to **(b) 2.14**.

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