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Two elements A and B with atomic numbers ZA and ZB are used to produce characteristic x–rays with frequencies A and B respectively. If ZA : ZB= 1 : 2, then vA : vB will be
  • a)
    1 :√2
  • b)
    1 : 8
  • c)
    4 : 1
  • d)
    1 : 4
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Two elements A and B with atomic numbers ZA and ZB are used to produce...
√v = a(z - b) , Ignoring screening effect (i.e. b=0)
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Two elements A and B with atomic numbers ZA and ZB are used to produce...
Understanding Characteristic X-Rays
Characteristic x-rays are emitted when electrons transition between energy levels in an atom. The frequency of emitted x-rays is directly related to the atomic number of the element.
Relation Between Atomic Number and Frequency
The frequency of the characteristic x-rays produced by an element is proportional to the square of its atomic number (Z). This relationship can be expressed as follows:
- v ∝ Z²
Given Condition
In this problem, we have two elements A and B with atomic numbers:
- ZA : ZB = 1 : 2
This means:
- ZA = k (for some constant k)
- ZB = 2k
Calculating Frequencies
Using the relationship v ∝ Z², we can express the frequencies:
- vA ∝ (ZA)² = (k)² = k²
- vB ∝ (ZB)² = (2k)² = 4k²
Now, we can find the ratio of frequencies vA to vB:
- vA : vB = k² : 4k²
This simplifies to:
- vA : vB = 1 : 4
Conclusion
Thus, the correct answer for the ratio of frequencies vA : vB is:
- 1 : 4 (Option D)
This demonstrates that as the atomic number increases, the frequency of the emitted x-rays increases significantly due to the square relationship.
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