Question

A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three αparticles in succession.The group of the periodic table to which the resulting daughter element would belong is:

• A. Group 14
• B. Group 16
• C. Group 4
• D. Group 6

Answer 1

The Correct Option is A

To solve this problem, we need to understand the process of radioactive decay involving α-particle emission and its effect on the periodic table position of an element.

Concept Explanation:

An α-particle consists of 2 protons and 2 neutrons, which is equivalent to a helium nucleus. When a nucleus emits an α-particle, its atomic number decreases by 2 and its mass number decreases by 4.

The question states that a nuclide of an alkaline earth metal undergoes three successive α-particle emissions. Let's analyze this step-by-step:

1. Alkaline earth metals belong to Group 2 of the periodic table. Common examples include Beryllium (Be), Magnesium (Mg), Calcium (Ca), etc.
2. Assume a general nuclide of an alkaline earth metal with atomic number Z. After the emission of one α-particle, its atomic number becomes Z - 2.
3. After three α-particle emissions, the cumulative decrease in atomic number will be (3 \times 2 = 6).
4. Thus, the new atomic number of the nuclide will be Z - 6.
5. An alkaline earth metal initially belongs to Group 2. After the reduction in the atomic number by 6, the resulting element will be in the group corresponding to (2 + 6 = 8) atomic numbers higher.

The original metal was in Group 2. Reducing the atomic number by 6 places it 6 groups higher. Thus, the final group in which the daughter element will belong will be Group 14 of the periodic table.

Conclusion:

Therefore, after the decay by emission of three α-particles in succession, the resulting daughter element will belong to Group 14.

Hence, the correct answer is Group 14.