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:
What is the major X-ray emitted by the sample?

Options

A. `K_(papha)`
B. `K_(beta)`
C. `K_(gamma)`
D. `K_(delta)`

Explanation

To determine the major X-ray emitted by the sample, we will analyze the transitions that can occur in the given scenario involving the `Li^(+ +)` ion and the other metallic atom with four orbits.

### Step-by-Step Solution:

1. **Understanding the Electron Configuration**:
- The `Li^(+ +)` ion has one electron in a very high energy level, denoted as the nth shell, where n is large.
- The other metallic atom has four orbits, which we can label as n=1 (K shell), n=2 (L shell), n=3 (M shell), and n=4 (N shell).

2. **Identifying Possible Transitions**:
- When an electron transitions from a higher energy level to a lower energy level, it emits energy in the form of X-rays.
- The possible transitions for the metallic atom with four orbits are:
- From n=4 to n=3
- From n=4 to n=2
- From n=4 to n=1
- From n=3 to n=2
- From n=3 to n=1
- From n=2 to n=1

3. **Classifying the X-ray Emissions**:
- The transitions can be classified as follows:
- **K-alpha (Kα)**: Transition from n=2 to n=1
- **K-beta (Kβ)**: Transition from n=3 to n=1
- **K-gamma (Kγ)**: Transition from n=4 to n=1
- **K-delta (Kδ)**: Transition from n=4 to n=2 (not typically classified as K-series)

4. **Determining the Major X-ray Emission**:
- The major X-ray emitted corresponds to the transition with the largest energy difference.
- The energy difference is greatest when transitioning from n=4 to n=1, which corresponds to K-gamma (Kγ).
- Therefore, the major X-ray emitted by the sample is K-gamma (Kγ).

5. **Final Answer**:
- The major X-ray emitted by the sample is **K-gamma (Kγ)**.