Question:
As a result of radioactive decay, \( ^{238}U \) is converted into \( ^{234}Pa \). The particles emitted during this decay are:
As a result of radioactive decay, \( ^{238}U \) is converted into \( ^{234}Pa \). The particles emitted during this decay are:
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In alpha decay, the nucleus loses two protons and two neutrons. In beta decay, a neutron converts into a proton, and an electron is emitted.
Updated On: Apr 22, 2026
- a proton and a neutron
- a proton and two \( \alpha \)-particles
- an \( \alpha \)-particle and a \( \beta \)-particle
- two \( \beta \)-particles and a proton
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The Correct Option is C
Solution and Explanation
Step 1: Understand the decay process.
When Uranium-238 (\( ^{238}U \)) undergoes radioactive decay, it transforms into Protactinium-234 (\( ^{234}Pa \)) through a two-step process. The decay of \( ^{238}U \) involves the emission of an \( \alpha \)-particle, followed by the emission of a \( \beta \)-particle.
Step 2: Explanation of emissions.
- The first step involves the emission of an \( \alpha \)-particle, which is a helium nucleus (\( ^4_2He \)), causing a decrease in the atomic number by 2 and the mass number by 4. - The second step involves the emission of a \( \beta \)-particle, which is an electron, resulting in the conversion of a neutron into a proton.
Step 3: Conclusion.
Thus, the correct emission products in this case are an \( \alpha \)-particle and a \( \beta \)-particle.
When Uranium-238 (\( ^{238}U \)) undergoes radioactive decay, it transforms into Protactinium-234 (\( ^{234}Pa \)) through a two-step process. The decay of \( ^{238}U \) involves the emission of an \( \alpha \)-particle, followed by the emission of a \( \beta \)-particle.
Step 2: Explanation of emissions.
- The first step involves the emission of an \( \alpha \)-particle, which is a helium nucleus (\( ^4_2He \)), causing a decrease in the atomic number by 2 and the mass number by 4. - The second step involves the emission of a \( \beta \)-particle, which is an electron, resulting in the conversion of a neutron into a proton.
Step 3: Conclusion.
Thus, the correct emission products in this case are an \( \alpha \)-particle and a \( \beta \)-particle.
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