If the frequency of `K_(alpha), K_(beta)` and `L_(alpha)` , X-ray lines of a substance are `v_(K_(alpha)), v_(K_(beta))`, and `v_(L_(beta))`

Options

A. `v_(K_(beta))=v_(K_(alpha))+v_(L_(alpha))`
B. `v_(K_(beta))=v_(K_(alpha))-v_(L_(alpha))`
C. `v_(K_(beta))=(v_(K_(alpha)))/(v_(L_(alpha)))`
D. `v_(K_(beta))^(2)=(v_(K_(alpha))*v_(L_(alpha)))/(2)`

Explanation

To solve the problem regarding the frequencies of the Kα, Kβ, and Lα X-ray lines of a substance, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Energy Levels**:
- The K and L series correspond to different energy levels of electrons in an atom. The K series refers to transitions involving the innermost shell (n=1), while the L series refers to transitions involving the second shell (n=2).

2. **Identifying the Transitions**:
- Kα transition occurs when an electron falls from the L shell (n=2) to the K shell (n=1).
- Kβ transition occurs when an electron falls from the M shell (n=3) to the K shell (n=1).
- Lα transition occurs when an electron falls from the M shell (n=3) to the L shell (n=2).

3. **Using the Energy-Frequency Relation**:
- The energy of a photon emitted during these transitions can be expressed as:
\[ E = h \nu \]
where \(E\) is the energy, \(h\) is Planck's constant, and \(\nu\) is the frequency.

4. **Setting Up the Energy Equations**:
- For Kβ, the energy can be expressed as:
\[ E_{K\beta} = E_{K\alpha} + E_{L\alpha} \]
- In terms of frequency, this becomes:
\[ h \nu_{K\beta} = h \nu_{K\alpha} + h \nu_{L\alpha} \]

5. **Cancelling Planck's Constant**:
- Since \(h\) is a common factor, we can cancel it out from both sides:
\[ \nu_{K\beta} = \nu_{K\alpha} + \nu_{L\alpha} \]

6. **Conclusion**:
- The correct relation among the frequencies is:
\[ \nu_{K\beta} = \nu_{K\alpha} + \nu_{L\alpha} \]
- Therefore, the correct option is A.