Problem 110 - A Complete Resource Book in Chemistry for JEE Main

Question

\({ }_{90} \mathrm{Th}^{232}\) decays to \({ }_{82} \mathrm{~Pb}^{206} .\) How many \(\alpha\) and \(\beta\) particles are emitted? (a) \(7 \alpha, 6 \beta\) (b) \(6 \alpha, 7 \beta\) (c) \(4 \alpha, 3 \beta\) (d) none of these

Step-by-Step Solution

Answer

6 \(\alpha\) and 4 \(\beta\) particles are emitted; answer is (d) none of these.

Step 1: Understand the Basics

A decay from \(A_{Z}X^A\) to \(A'_{Z'}Y^{A'}\) involves a change in atomic number (\(Z\)) and mass number (\(A\)). Each \(\alpha\)-particle decay reduces the atomic number by 2 and the mass number by 4, while each \(\beta\)-particle decay increases the atomic number by 1.

Step 2: Analyze Given Details

The decay starts from \(^{232}_{90}\mathrm{Th}\) and ends at \(^{206}_{82}\mathrm{Pb}\). The initial mass number \( (A) \) is 232 and the final mass number \( (A') \) is 206; the initial atomic number \( (Z) \) is 90 and the final atomic number \( (Z') \) is 82.

Step 3: Calculate Change in Mass Number

The change in mass number \( (A - A') \) is calculated as \(232 - 206 = 26\). Since each \(\alpha\)-particle decay reduces the mass number by 4, divide 26 by 4 to find the number of \(\alpha\) particles: \(\frac{26}{4} = 6.5\). Round down to 6 (since we can only have whole numbers of particles), indicating 6 \(\alpha\) particles are emitted.

Step 4: Calculate Change in Atomic Number

For \(\alpha\)-particle decays, the atomic number drops by 2 for each particle, giving a total decrease of \(6 \, \alpha \times 2 = 12\) in atomic number. The starting atomic number is 90, so after 6 \(\alpha\) decays, the atomic number is \(90 - 12 = 78\). To reach the final atomic number 82, we need to calculate the additional increase caused by \(\beta\)-particles: \(82 - 78 = 4\). Thus, 4 \(\beta\) particles are needed.

Step 5: Conclusion

The decay process involves the emission of 6 \(\alpha\) and 4 \(\beta\) particles. Since this matches none of the options exactly, the answer is (d) none of these.

Key Concepts

Alpha Decay

Alpha decay is a type of nuclear decay where an unstable nucleus emits an alpha particle to become more stable. An alpha particle consists of 2 protons and 2 neutrons, identical to a helium nucleus. This process results in a decrease in:

• Mass number by 4
• Atomic number by 2

This happens because the nucleus is losing 2 protons (which define the element) and 2 neutrons. An example of alpha decay is when Thorium-232 ( ^{232}_{90} ext{Th}) decays to Lead-206 ( ^{206}_{82} ext{Pb}). Here, ^{90}Th emits 6 alpha particles, each reducing the mass number by 4 (total 24) and the atomic number by 12.

Beta Decay

Beta decay occurs when a nucleus emits a beta particle, which is either an electron or a positron. This decay does not change the mass number but alters the atomic number:

• +1 for beta-minus decay (electron emission)
• -1 for beta-plus decay (positron emission)

During beta-minus decay, a neutron is converted to a proton, increasing the atomic number by 1. For beta-plus decay, the opposite occurs. In our example, to reach the atomic number 82 from 78 after 6 alpha decays, the process involves the emission of 4 beta-minus particles, increasing the atomic number by 4.

Mass Number

The mass number of an atom is the total count of protons and neutrons in its nucleus. It's denoted by the letter A. This number helps identify isotopes of elements and does not influence chemical properties, which are dictated by electrons. During nuclear decay:

• Alpha decay reduces the mass number by 4 per alpha particle
• Beta decay does not change the mass number

In our decay example, the mass number changes from 232 to 206, a decrease caused entirely by the emission of 6 alpha particles (as each reduces it by 4, totaling 24). No change in mass number occurs from beta decay.

Atomic Number

The atomic number, represented by the letter Z, is the number of protons in an atom's nucleus. It defines the element and its position in the periodic table. Changes in atomic number occur during nuclear decay:

• Decreases by 2 per alpha particle emission
• Increases by 1 per beta-minus particle emission

For the decay from ^{232}_{90} ext{Th} to ^{206}_{82} ext{Pb}, the atomic number starts at 90 and ends at 82. Initially, 6 alpha particles reduce this number to 78, and 4 beta particles bring it back up to 82.