The electron in a `Li^(+ +)` ion is the nth shell , `n` being very large. One of the K-electron in another matallic atom has been knocked out. The second matel has four orbits. Now, we take two sample one of `Li^(+ +)` ion and the other of the second mateillic ions. Suppose the probability of electronic transition from higher to lower energy level is directly proportional to the energy difference between the two shells. Take `hc = 1224 eV nm` , where `h` is Planck's constant and `c` the velocity of light in vacuum. It is found that major electromagnetic wavws emitted from the two sample are identical. Now , answer the following questions:
The wavelength of this major X-ray is

Options

A. `0.90 Å`
B. `1.0 Å`
C. `1.1 Å`
D. none of these

Explanation

To solve the problem, we need to find the wavelength of the major X-ray emitted from the `Li^(+ +)` ion and the second metallic ion after an electron transition. Let's break down the solution step by step.

### Step 1: Understanding the Energy Levels
For a hydrogen-like atom, the energy levels can be calculated using the formula:

\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]

where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 2: Calculate Energy Levels for `Li^(+ +)`
The atomic number \( Z \) for lithium is 3. We need to calculate the energy for the first four energy levels (n=1 to n=4).

1. **For n = 1:**
\[ E_1 = -\frac{13.6 \cdot 3^2}{1^2} = -\frac{13.6 \cdot 9}{1} = -122.4 \, \text{eV} \]

2. **For n = 2:**
\[ E_2 = -\frac{13.6 \cdot 3^2}{2^2} = -\frac{13.6 \cdot 9}{4} = -30.6 \, \text{eV} \]

3. **For n = 3:**
\[ E_3 = -\frac{13.6 \cdot 3^2}{3^2} = -\frac{13.6 \cdot 9}{9} = -13.6 \, \text{eV} \]

4. **For n = 4:**
\[ E_4 = -\frac{13.6 \cdot 3^2}{4^2} = -\frac{13.6 \cdot 9}{16} = -7.65 \, \text{eV} \]

### Step 3: Calculate the Energy Difference
Now, we need to find the energy difference between the levels involved in the transition. The transition is from \( n = 4 \) to \( n = 1 \):

\[ \Delta E = E_4 - E_1 = (-7.65) - (-122.4) = 114.75 \, \text{eV} \]

### Step 4: Calculate the Wavelength
Using the relationship between energy and wavelength:

\[ E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E} \]

Given \( hc = 1224 \, \text{eV nm} \):

\[ \lambda = \frac{1224 \, \text{eV nm}}{114.75 \, \text{eV}} \approx 10.66 \, \text{nm} \]

### Step 5: Convert to Angstroms
To convert nanometers to angstroms (1 nm = 10 angstroms):

\[ \lambda = 10.66 \, \text{nm} = 106.6 \, \text{angstroms} \]

### Final Answer
The wavelength of the major X-ray is approximately:

\[ \lambda \approx 10.66 \, \text{nm} \quad \text{or} \quad 106.6 \, \text{angstroms} \]

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