Problem 26 - A Complete Resource Book in Chemistry for JEE Main

Question

In successive emission of \(\beta\) and \(\alpha\) particles, how many \(\alpha\) and \(\beta\) particles should be emitted for the natural \(\left(4 n+1\right.\) series) conversion of \(_{94} P u^{241}\) to \({ }_{92} U^{233} ?\) (a) \(2 \alpha, \beta\) (b) \(2 \alpha, 2 \beta\) (c) \(2 \alpha, 3 \beta\) (d) \(\alpha, 2 \beta\)

Step-by-Step Solution

Answer

The answer is (b) 2 alpha, 2 beta.

Step 1: Write the starting and ending nuclei

In this problem, we need to find out how many alpha and beta particles are emitted during the transformation of \(_{94}^{241}Pu\) to \(_{92}^{233}U\). Here, \(_{94}^{241}Pu\) is the starting nucleus, and \(_{92}^{233}U\) is the ending nucleus.

Step 2: Determine the change in atomic number (Z) and mass number (A)

The starting nucleus \(_{94}Pu^{241}\), has an atomic number of 94 and a mass number of 241. The ending nucleus \(_{92}U^{233}\), has an atomic number of 92 and a mass number of 233. Calculate the difference in atomic number \(\Delta Z = 94 - 92 = 2\) and the difference in mass number \(\Delta A = 241 - 233 = 8\).

Step 3: Calculate the number of alpha particles needed

Each alpha particle reduces the mass number by 4. We need to reduce the total mass number by 8, so the number of alpha particles required is \(\frac{8}{4} = 2\). Each alpha emission also decreases the atomic number by 2. Thus, after 2 alpha emissions, the atomic number decreases by 4.

Step 4: Calculate the number of beta particles needed

Each beta particle increases the atomic number by 1. As the atomic number needs to be decreased by only 2 to reach 92 from the initial 94, but also needs to account for the change from the two alpha particle emissions \((94 - 4 = 90)\), we need additional \(2\) beta emissions. Therefore, the number of beta particles needed is also 2, giving an atomic number of \((90 + 2) = 92\).

Step 5: Verify the calculation

Finally, verify the solution by considering: 2 alpha emissions (reduce mass by 8 and atomic number by 4) and 2 beta emissions (adjust atomic number by +2 from 90 back to 92). Both mass and atomic numbers check out correctly with the transformation to \(_{92}U^{233}\).

Key Concepts

Alpha Decay

Alpha decay is a common form of radioactive decay where an unstable nucleus emits an alpha particle.
An alpha particle consists of 2 protons and 2 neutrons, which is the same as a helium nucleus.

• This emission leads to a decrease in both the atomic number and the mass number of the original nucleus.
• The atomic number decreases by 2 because two protons are lost.
• The mass number decreases by 4 as both two protons and two neutrons—adding to four nucleons—are lost.

To illustrate, in the transformation of \(_{94}^{241}Pu\) to \(_{92}^{233}U\), alpha particles are emitted. Each alpha decay directly reduces \(A\) and \(Z\).
After emitting 2 alpha particles, the mass number drops from 241 to 233 and the atomic number reduces from 94 to a transitional 90. This sets the stage for other decay processes.

Beta Decay

Beta decay occurs when a beta particle is emitted from the nucleus, typically when a neutron is transformed into a proton.
The emitted particle is a high-energy electron or positron, known as  beta minus ( 1) decay or beta plus (1) decay, respectively.
During beta minus decay:

• The atomic number increases by 1 because a neutron converts to a proton.
• The mass number remains unchanged because no nucleons are lost or gained.

In the described exercise, following two alpha decays, two beta decays are necessary to adjust the atomic number back up to 92. This compensates for the initial drop to 90 due to the two alpha emissions.

Atomic Number

The atomic number, represented as \(Z\), is crucial for identifying an element.

• It defines the number of protons in an atom's nucleus, which in turn determines the element's identity and position on the periodic table.
• In nuclear reactions, changes in the atomic number signify a transformation into a different element.

For example, when \(_{94}Pu\) undergoes decay processes, the net change in its atomic number is calculated to ensure the transformation into the desired element, \(_{92}U\).
During this process, the atomic number changes through alpha and beta decay, ultimately reaching the target end point.

Mass Number

Mass Number, noted as \(A\), is a count of the total protons and neutrons in a nucleus.

• While the atomic number determines the element's identity, the mass number reveals the specific isotope of that element.
• Because an alpha particle emission involves the loss of 4 nucleons, it directly affects the nucleus's mass number.

As seen in our problem, two alpha decays reduce the mass number from 241 to 233.
This essential calculation allows us to track the changes in isotopic composition through the steps of the nuclear reaction process.