Question

\({ }_{2} \mathrm{U}^{238}\) is radioactive and it emits \(\alpha\) and \(\beta\) particles to form \({ }_{82} \mathrm{~Pb}^{206} .\) The number of \(\alpha\) and \(\beta\) particles are respectively (a) 8,6 (b) 6,8 (c) 4,2 (d) 8,4

Short answer

(a) 8, 6

Step-by-step solution

Understand Alpha and Beta Decay

An \(\alpha\)-particle is a helium nucleus consisting of 2 protons and 2 neutrons, resulting in the loss of 4 atomic mass units (AMU) and a reduction of 2 in atomic number. A \(\beta\)-particle is an electron or positron emitted during the transformation of a neutron to a proton (or vice versa), which affects the atomic number by +1 or -1 but has no significant impact on the AMU.

Determine the Mass Change Due to Alpha Decay

The original uranium nucleus, \( _{92}^{238} \text{U} \), changes to \( _{82}^{206} \text{Pb} \). First, calculate the change in mass: \( 238 - 206 = 32 \). Since each \(\alpha\)-particle reduces the mass by 4, we calculate the number of \(\alpha\)-decays. Hence, \[ n = \frac{32}{4} = 8 \].This implies there were 8 \(\alpha\)-particles emitted.

Calculate Effect of Alpha Particle on Atomic Number

Each \(\alpha\)-particle decreases the atomic number by 2. With 8 \(\alpha\)-particles emitted, the atomic number decreases by:\[ 8 \times 2 = 16 \]. So, the new atomic number is \[ 92 - 16 = 76 \].

Determine Beta Particles Required to Reach Atomic Number of Lead

The final product, lead \( _{82}^{206} \text{Pb} \), has an atomic number of 82. From uranium (92) through alpha decay, we reached the atomic number 76. To achieve 82, we need an increase of:\[ 82 - 76 = 6 \].Each \(\beta\)-decay increases the atomic number by 1, so 6 \(\beta\)-particles are needed.

Verify the Calculation

Verify that the decay sequence correctly results in \( _{82}^{206} \text{Pb} \) with 8 \(\alpha\)-particles (mass number reduction) accounting for the mass shift from 238 to 206, and 6 \(\beta\)-particles accounting for the increase from atomic number 76 to 82.

Key concepts

Alpha Decay

Alpha decay is a process where an atomic nucleus emits an alpha particle, consisting of 2 protons and 2 neutrons, which is essentially a helium nucleus. This emission results in the reduction of the atomic mass of the original element by 4 units and the atomic number by 2, shifting the element two places back in the periodic table.
When uranium \(_{92}^{238} \text{U}\) undergoes alpha decay, it emits an alpha particle and changes into a new element with a lower atomic number and mass. If multiple alpha particles are emitted, each one reduces the mass number by 4 and the atomic number by 2, as seen in the transition of uranium to lead.
In our example, the mass number decreased by 32 due to the emission of 8 alpha particles, as calculated by dividing the total mass reduction (32) by the mass of one alpha particle (4).

Beta Decay

Beta decay involves the transformation of a neutron into a proton, resulting in the emission of a beta particle, which could be an electron or a positron. Unlike alpha decay, beta decay doesn't significantly affect the atomic mass but increases or decreases the atomic number by one, depending on whether a beta-minus or beta-plus particle is emitted.
In our scenario, following the alpha decay, uranium eventually turns into lead by emitting beta particles. The emission of these beta particles increases the atomic number without changing the overall mass, critical for matching the final atomic number of lead. In our example, 6 beta particles were necessary to increase the atomic number from 76 (after alpha decay) to 82 (lead).

• Each beta particle increases the atomic number by 1, indicating a contribution to the final atomic identity without altering the mass number.

Uranium to Lead Conversion

The conversion of uranium to lead via radioactive decay is a multi-step process involving both alpha and beta decay. Initially, uranium \(_{92}^{238} \text{U}\) undergoes a series of alpha decays, reducing both its atomic mass and atomic number significantly.
Following alpha decay, uranium is not directly converted to lead, since the atomic number is still not aligned with lead's number of 82. Beta decay then plays a crucial role in ensuring that the atomic number is adjusted appropriately while maintaining the lowered mass. This combination of decays allows uranium to finally stabilize as lead.

• The sequence involves emitting sufficient alpha particles to adjust the mass, then enough beta particles to adjust the atomic number.

Atomic Mass Unit Reduction

In radioactive decay, the atomic mass unit (AMU) reduction is primarily influenced by alpha decay. Each alpha particle emitted results in a reduction of the element's atomic mass by 4 AMU.
In our specific example, the uranium's atomic mass was reduced from 238 to 206 through the emission of 8 alpha particles. To determine the mass reduction through decay, the original mass number is subtracted by the final mass number, giving the total reduction. Dividing this reduction by the AMU loss per alpha particle (4) confirms the number of alpha decays required to achieve the observed transformation.
This understanding is crucial in calculating decay sequences and understanding the behavior of heavy elements undergoing radioactive decay.