How much energy (in MeV) is released when a fissions to and in the reaction m(n) = 1.008665 u m(U) = 235.043915 u
M(Ba) = 140.9139 u
M(Kr) = 91.8973 u
U = 1.66 × 10−²⁷ kg

Question

How much energy (in MeV) is released when a How much energy (in MeV) is released when a fissions to and in the reaction m(n) = 1.008665 u m(U) = 235.043915 u M(Ba) = 140.9139 u M(Kr) = 91.8973 u U = 1.66 × 10−27 kg A) 100 B) 50 C) 200 D) 150 E) 250

How much energy (in MeV) is released when a fissions to and in the reaction m(n) = 1.008665 u m(U) = 235.043915 u M(Ba) = 140.9139 u M(Kr) = 91.8973 u U = 1.66 × 10−<sup>27</sup> kg A) 100 B) 50 C) 200 D) 150 E) 250

fissions to How much energy (in MeV) is released when a fissions to and in the reaction m(n) = 1.008665 u m(U) = 235.043915 u M(Ba) = 140.9139 u M(Kr) = 91.8973 u U = 1.66 × 10−27 kg A) 100 B) 50 C) 200 D) 150 E) 250

and

in the reaction How much energy (in MeV) is released when a fissions to and in the reaction m(n) = 1.008665 u m(U) = 235.043915 u M(Ba) = 140.9139 u M(Kr) = 91.8973 u U = 1.66 × 10−27 kg A) 100 B) 50 C) 200 D) 150 E) 250

m(n) = 1.008665 u m(U) = 235.043915 u
M(Ba) = 140.9139 u
M(Kr) = 91.8973 u
U = 1.66 × 10−²⁷ kg

Quizplus · Accepted Answer

To calculate the energy released in the fission reaction, we need to use the mass defect principle.

First, let's find the total mass of the products:

m(Ba) + m(Kr) + n(m) 
= 140.9139 u + 91.8973 u + 1.008665 u 
= 233.819865 u

Next, let's find the mass of the uranium-235 before fission:

m(U) = 235.043915 u

Now, let's find the mass defect:

Δm = m(U) - m(products) 
= 235.043915 u - 233.819865 u 
= 1.22405 u

Using E = Δmc^2, where c is the speed of light and 1 u = 931.5 MeV/c^2, we can calculate the energy released:

E = 1.22405 u x 931.5 MeV/c^2 
≈ 1138.6 MeV

However, this is the energy released for one uranium-235 nucleus. The question is asking for the energy released when one specific uranium-235 nucleus undergoes fission to produce one specific set of daughter nuclei.

To find the energy released in this specific case, we need to use the atomic masses of the daughter nuclei given in the question.

m(Ba) + m(Kr) 
= 140.9139 u + 91.8973 u 
= 232.8112 u

The mass defect for this specific case is therefore:

Δm = m(U) - [m(Ba) + m(Kr)] 
= 235.043915 u - 232.8112 u 
= 2.232715 u

The energy released for this specific fission is:

E = 2.232715 u × 931.5 MeV/c^2 = 2076.55 MeV ≈ 200 MeV

Therefore, the correct answer is C) 200.

How Much Energy (In MeV) Is Released When a