The easiest fusion reaction to initiate is 2₁H + ³H → ₂He + 'on Calculate the energy released, in kJ, per nucleus of 42He produced. The masses of the relevant particles are as follows (1 amu = 1.66×10-27 kg): Particle Mass (amu) H 2.01410 зн 3.01605 He 4.00260 1.00866 5.4858 x 10-4 In ie
The easiest fusion reaction to initiate is 2₁H + ³H → ₂He + 'on Calculate the energy released, in kJ, per nucleus of 42He produced. The masses of the relevant particles are as follows (1 amu = 1.66×10-27 kg): Particle Mass (amu) H 2.01410 зн 3.01605 He 4.00260 1.00866 5.4858 x 10-4 In ie
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter21: Nuclear Chemistry
Section: Chapter Questions
Problem 21.51QE
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![**Fusion Reaction Energy Calculation**
The easiest fusion reaction to initiate is:
\[ ^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n} \]
To calculate the energy released, in kJ, per nucleus of \( ^4_2\text{He} \) produced, we must consider the masses of the relevant particles. The masses of these particles in atomic mass units (amu) are provided as follows:
\[ 1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg} \]
| **Particle** | **Mass (amu)** |
|--------------|----------------|
| \( ^2_1\text{H} \) | 2.01410 |
| \( ^3_1\text{H} \) | 3.01605 |
| \( ^4_2\text{He} \) | 4.00260 |
| \( ^1_0\text{n} \) | 1.00866 |
| \( ^0_{-1}\text{e} \) | 5.4858 x 10^-4 |
By using these masses, the energy released in the fusion reaction can be calculated. For a detailed explanation and step-by-step process, refer to the subsequent sections.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74446e5c-3f60-43cb-b67a-6a64a4e1e483%2Ff0713184-6fb5-4146-b88b-ed5f43a5e04f%2Fxedizf8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Fusion Reaction Energy Calculation**
The easiest fusion reaction to initiate is:
\[ ^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n} \]
To calculate the energy released, in kJ, per nucleus of \( ^4_2\text{He} \) produced, we must consider the masses of the relevant particles. The masses of these particles in atomic mass units (amu) are provided as follows:
\[ 1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg} \]
| **Particle** | **Mass (amu)** |
|--------------|----------------|
| \( ^2_1\text{H} \) | 2.01410 |
| \( ^3_1\text{H} \) | 3.01605 |
| \( ^4_2\text{He} \) | 4.00260 |
| \( ^1_0\text{n} \) | 1.00866 |
| \( ^0_{-1}\text{e} \) | 5.4858 x 10^-4 |
By using these masses, the energy released in the fusion reaction can be calculated. For a detailed explanation and step-by-step process, refer to the subsequent sections.
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