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 X-ray having least intensity emitted by the second sample?

Options

A. `K_(papha)`
B. `L_(papha)`
C. `M_(papha)`
D. data insufficient

Explanation

To solve the problem, we need to analyze the electronic transitions in the given systems and identify the X-ray with the least intensity emitted by the second sample.

### Step-by-Step Solution:

1. **Understanding the Systems**:
- We have a `Li^(+ +)` ion with an electron in the nth shell, where n is very large.
- The second metallic atom has four orbits (shells), which we can label as n=1, n=2, n=3, and n=4.

2. **Identifying the Transition**:
- The question states that the probability of electronic transition from a higher to a lower energy level is directly proportional to the energy difference between the two shells.
- The least intense X-ray emission corresponds to the transition that involves the smallest energy difference.

3. **Energy Level Transitions**:
- For the second metallic atom, the transitions can occur from:
- n=2 to n=1 (Kα transition)
- n=3 to n=2
- n=4 to n=3
- The transition from n=2 to n=1 will have the least energy difference, as it is the closest pair of energy levels.

4. **Calculating the Energy Difference**:
- The energy difference for a transition from n=2 to n=1 can be calculated using the formula for the energy levels of hydrogen-like atoms:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
- For n=1:
\[ E_1 = -\frac{Z^2 \cdot 13.6}{1^2} = -Z^2 \cdot 13.6 \, \text{eV} \]
- For n=2:
\[ E_2 = -\frac{Z^2 \cdot 13.6}{2^2} = -\frac{Z^2 \cdot 13.6}{4} \, \text{eV} \]
- The energy difference (ΔE) for the transition from n=2 to n=1 is:
\[ \Delta E = E_1 - E_2 = -Z^2 \cdot 13.6 + \frac{Z^2 \cdot 13.6}{4} = Z^2 \cdot 13.6 \left(1 - \frac{1}{4}\right) = Z^2 \cdot 13.6 \cdot \frac{3}{4} \]

5. **Identifying the X-ray Emission**:
- The transition from n=2 to n=1 corresponds to the Kα line in X-ray emissions.
- Since this transition has the least energy difference, it will emit the X-ray with the least intensity.

6. **Conclusion**:
- Therefore, the X-ray having the least intensity emitted by the second sample is the Kα line.

### Final Answer:
The X-ray having the least intensity emitted by the second sample is the Kα line.