Question

We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the atoms and any other atomic particles that are part of the reaction. With this in mind, calculate the mass defect in amu and kg, and binding energy in J for the reaction of 1 neutron with U-235 to produce Te-137, Zr-97, and some neutrons. Balance the nuclear equation first before proceeding.
U-235 = 235.04393 amu, 01n = 1.00867 amu, Te-137 = 136.92532 amu, Zr-97 = 96.91095 amu

Boron-10 undergoes fission via thermal neutron capture of a single neutron to produce lithium-7, an alpha particle, and energy. Write a balanced nuclear equation, and then calculate the mass defect in amu and kg, and binding energy in J.
B-10 = 10.01294 amu, 01n = 1.00867 amu, Li-7 = 7.01600 amu, α (He-4) = 4.00260 amu

Answer 1

Answer:

See Explanation

Explanation:

$\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n$

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) $\frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy$

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J