Consider the following: 13N -> 13C + β+ + νe Calculate the energy released (in MeV) in the β+ decay of 13N, the equation for which is given above. The masses of 13N and 13C are 13.005738 and 13.003355 u, respectively. (Assume 1 u = 931.5 MeV/c^2.)
Added by Tiffany M.
Step 1
- Mass of 13N = 13.005738 u - Mass of 13C = 13.003355 u - Mass difference (Δm) = Mass of 13N - Mass of 13C - Δm = 13.005738 u - 13.003355 u = 0.002383 u Show more…
Show all steps
Close
Your feedback will help us improve your experience
Maitreya E and 84 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Calculate the energy (in MeV) released in the α decay of 226Ra. (Assume 1 u = 931.5 MeV/c2.) ______MeV (b) What fraction of the mass of a single 226Ra is destroyed in the decay? The mass of 222Rn is 222.01757 u.
Shaiju T.
Consider the induced nuclear reaction ${ }_{1}^{2} \mathrm{H}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He} .$ The atomic masses are ${ }_{1}^{2} \mathrm{H}(2.014102 \mathrm{u}),{ }_{7}^{14} \mathrm{~N}(14.003074 \mathrm{u}),{ }_{6}^{12} \mathrm{C}(12.000000 \mathrm{u}),$ and ${ }_{2}^{4} \mathrm{He}$ $(4.002603 \mathrm{u}) .$ Determine the energy (in $\mathrm{MeV}$ ) released when the ${ }_{6}^{12} \mathrm{C}$ and ${ }_{2}^{4} \mathrm{He}$ nuclei are formed in this manner.
Calculate the energy (in MeV) released in the triplealpha process 3 $^4He \rightarrow ^{12}C$.
Particle Physics and Cosmology
The Beginning of Time
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD