Question

The naturally occurring radioactive decay series that begins with \({ }_{92}^{235} \mathrm{U}\) stops with formation of the stable \({ }_{82}^{207} \mathrm{~Pb}\) nucleus. The decays proceed through a series of alpha-particle and beta-particle emissions. How many of each type of emission are involved in this series?

Short answer

In the decay series of Uranium-235 (\(^{235}_{92} \mathrm{U}\)) to Lead-207 (\(^{207}_{82} \mathrm{Pb}\)), there are 7 alpha decays and 4 beta decays.

Step-by-step solution

Understand Alpha and Beta Decay

During an alpha decay, an unstable nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. This results in a decrease in the atomic number (Z) by 2 and a decrease in the mass number (A) by 4. During a beta decay, a neutron in the nucleus gets converted into a proton, and an electron (beta particle) is emitted. This results in an increase in the atomic number (Z) by 1, while the mass number (A) remains unchanged. Now that we understand how the atomic and mass numbers change during alpha and beta decays, we can proceed to determine the number of alpha and beta particle emissions in the given decay series.

Calculate the change in Atomic Number and Mass Number

The initial atomic number (Z1) and mass number (A1) of Uranium-235 are 92 and 235, respectively. The final atomic number (Z2) and mass number (A2) of Lead-207 are 82 and 207, respectively. The change in atomic number ΔZ = Z2 - Z1 = 82 - 92 = -10 The change in mass number ΔA = A2 - A1 = 207 - 235 = -28

Determine the number of Alpha and Beta emissions

Let 'a' be the number of alpha decays and 'b' be the number of beta decays. From Step 2, we know that after all the decays, the change in atomic number is -10 and change in mass number is -28. Since alpha decay decreases the atomic number by 2 and beta decay increases the atomic number by 1, we can write the equation: -2a + b = -10 Alpha decay also decreases the mass number by 4, while beta decay doesn't change the mass number. Hence, the equation becomes: -4a = -28 Now, we can solve the system of equations to find the values of 'a' and 'b'.

Solve the system of equations

From the second equation, we can find the number of alpha decays: -4a = -28 a = 7 Substitute the value of 'a' in the first equation: -2(7) + b = -10 -14 + b = -10 b = 4 Therefore, there are 7 alpha decays (a = 7) and 4 beta decays (b = 4) in the decay series of Uranium-235 to Lead-207.