Can anyone help me to calculate the energy released when 7Li and a deuteron (d) fuse to form 9Be,
Given only mass excess 7Li 16004 uu, d 14102 uu , and 9Be 12182 uu
uu might be microunit perhaps ???
Thank you explanation would be good thanks
To calculate the energy released when 7Li and a deuteron fuse to form 9Be, we need to use the concept of mass-energy equivalence expressed by Einstein's famous equation E = mc^2.
Here are the steps to calculate the energy released:
Step 1: Find the sum of the mass excess values for 7Li and d:
Mass excess 7Li = 16004 uu
Mass excess d = 14102 uu
Total mass excess = 16004 uu + 14102 uu = 30106 uu
Step 2: Find the mass of 9Be by subtracting the mass excess from the atomic mass:
Atomic mass of 9Be = 12182 uu
Mass of 9Be = Atomic mass - Mass excess = 12182 uu - 30106 uu = -17924 uu
Step 3: Convert the negative mass value to positive by taking the absolute value:
Abs(Mass of 9Be) = |-17924 uu| = 17924 uu
Step 4: Calculate the energy released using E = mc^2:
Energy released = (Mass of 9Be) * (Speed of light)^2 = (17924 uu) * (3 x 10^8 m/s)^2
Now, to convert the energy released into a more standard unit such as joules, we need to know the value of 1 uu (microunit) in kilograms. Unfortunately, without that information, we can't directly convert the energy into joules.
To determine the value of 1 uu in kilograms, it would be helpful to consult a reference source like a nuclear physics database or research paper, as the value could vary depending on the context of the calculations.
So, in conclusion, with the given information of mass excess for 7Li, d, and 9Be, we can calculate the energy released as described above. However, to convert it into joules, we would need the conversion factor for 1 uu to kilograms.