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 convert atomic mass units , uu, to kilograms see:
http://www.unitconversion.org/weight/kilograms-to-atomic-mass-units-conversion.html
Then use E = m c^2
thank you
Certainly! To calculate the energy released when 7Li and a deuteron (d) fuse to form 9Be, we can use Einstein's famous mass-energy equivalence equation: E = mc^2.
First, let's calculate the mass difference between the reactants (7Li and d) and the product (9Be). Mass excess is the mass of an atom minus the sum of the masses of its individual nucleons.
Mass excess of 7Li = 16004 uu
Mass excess of d = 14102 uu
Mass excess of 9Be = 12182 uu
Next, we can find the mass difference (∆m) between the reactants and the product:
∆m = (Mass excess of 7Li + Mass excess of d) - Mass excess of 9Be
∆m = (16004 uu + 14102 uu) - 12182 uu
= 30006 uu - 12182 uu
= 17824 uu
Now, using the mass difference (∆m), we can calculate the energy released (E) during the fusion reaction:
E = ∆m * c^2
Here, c represents the speed of light, which is approximately 3 x 10^8 meters per second.
E = 17824 uu * (3 x 10^8 m/s)^2
To convert uu (microunits) to kilograms, we need to know the conversion factor. Since uu was not defined in the question, we cannot proceed with a specific value.
If you have the conversion factor from uu to kilograms, you can substitute it into the equation to calculate the energy released. The result will be in joules (J).
If uu represents microunits (10^-6 units), you can convert it to kilograms by using the conversion factor 1 uu = 1.67 x 10^-27 kg. However, please note that uu is not a commonly used unit, so it is vital to ensure the accuracy of the conversion factor for your particular context.
Once you have the energy value in joules, you can convert it to other units if needed, such as electron volts (eV) or kilojoules (kJ), by using appropriate conversion factors.