(BONUS Extra Credit)
a. Write the [tex]$\beta+$[/tex] decay equation for [tex]${ }_6^{11} C$[/tex].
b. Calculate the energy released in the decay. The masses of [tex]${ }_6^{11} C$[/tex] is 11.011433 amu and [tex]${ }_5^{11} B$[/tex] 11.009305 amu. ([tex]$1 amu =1.66 \times 10^{-27} kg, E = mc ^2$[/tex] where [tex]$E =$[/tex] energy, [tex]$m =$[/tex] mass, and [tex]$c =$[/tex] speed of light [tex]$=3.0 \times 10^8 m / s$[/tex] )
Answer (2)
Write the β + decay equation: 6 11 C → 5 11 B + e + + n u e .
Calculate the mass difference: Δ m = m ( 6 11 C ) − m ( 5 11 B ) − m ( e + ) = 0.00157942 amu .
Convert the mass difference to kg: Δ m = 2.6218372 × 1 0 − 30 kg .
Calculate the energy released: E = Δ m c 2 = 1.4729 MeV .
E = 1.4729 MeV
Explanation
Problem Analysis We are given the mass of Carbon-11 ( 6 11 C ) as 11.011433 amu and the mass of Boron-11 ( 5 11 B ) as 11.009305 amu. We need to write the β + decay equation for 6 11 C and calculate the energy released in the decay. We will use the formula E = m c 2 to calculate the energy, where m is the mass difference between the initial and final states, and c is the speed of light.
Writing the Decay Equation The β + decay equation for 6 11 C is: 6 11 C → 5 11 B + e + + n u e where e + is a positron and n u e is an electron neutrino.
Calculating the Mass Difference To calculate the energy released, we first need to find the mass difference between the initial and final states. The initial state is 6 11 C , and the final state is 5 11 B + e + + n u e . The mass of the positron ( e + ) is approximately 0.00054858 amu. The mass of the electron neutrino is negligible.
The mass difference Δ m is given by: Δ m = m ( 6 11 C ) − m ( 5 11 B ) − m ( e + ) Δ m = 11.011433 amu − 11.009305 amu − 0.00054858 amu Δ m = 0.00157942 amu
Converting Mass Difference to kg Now, we convert the mass difference from amu to kg using the conversion factor 1 amu = 1.66 × 1 0 − 27 kg :
Δ m = 0.00157942 amu × 1.66 × 1 0 − 27 kg/amu Δ m = 2.6218372 × 1 0 − 30 kg
Calculating the Energy Released Next, we calculate the energy released using E = m c 2 , where c = 3.0 × 1 0 8 m/s :
E = ( 2.6218372 × 1 0 − 30 kg ) × ( 3.0 × 1 0 8 m/s ) 2 E = 2.35965348 × 1 0 − 13 J
Converting Energy to MeV Finally, we can convert the energy from Joules to MeV using the conversion factor 1 MeV = 1.602 × 1 0 − 13 J :
E = 1.602 × 1 0 − 13 J/MeV 2.35965348 × 1 0 − 13 J E ≈ 1.4729 MeV Therefore, the energy released in the decay is approximately 1.4729 MeV.
Final Answer The β + decay equation for 6 11 C is 6 11 C → 5 11 B + e + + n u e , and the energy released in the decay is approximately 1.4729 MeV.
Examples
Beta decay is a type of radioactive decay that is used in medical imaging, such as PET scans. In PET scans, a radioactive isotope that undergoes beta+ decay is injected into the body. The positrons emitted during the decay annihilate with electrons in the body, producing gamma rays that can be detected by a scanner. By analyzing the detected gamma rays, doctors can create images of the inside of the body and diagnose various medical conditions. The energy released during the decay affects the resolution and penetration depth of the scan, making accurate calculations crucial for effective imaging.
The β + decay equation for Carbon-11 is 6 11 C → 5 11 B + e + + ν e . The energy released in this decay is approximately 1.4729 MeV. This calculation involves determining the mass difference between the initial and final particles and then using Einstein's equation to find the energy equivalent.
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