1. The problem statement, all variables and given/known data (a) Calculate the disintegration energy when 232/92U decays by alpha emission into 228/90Th. Atomic masses of 232/92U and 228/90Th are 232.037156u and 228.028741u, respectively. (b) For the 232/92U decay in part (a), how much of the disintegration energy will be carried off by the alpha particle? Given: Mass of 4/2He = 4.002603u c^2 = 931.5MeV
2. Relevant equations E=mc^2
3. The attempt at a solution Well for part (a), first I found the difference in the starting masses and the end masses ie, 232.037156u - (228.028741u + 4.002603u) = 0.005812u I then put this into the equation and got 5.413878MeV. I thought this was right until I read part (b) and now I'm starting to think this might be how I'm meant to do that part, not part (a). Could anyone tell me if I'm even on the right track with this question or should I be using different equations?
If u put a cold & hot together, the temp with start to even out to the same temp.
Keep in mind that the angles of a triangle add up to 180°. The total angle measurement of a line segment is also 180°. The total angle measurement of a right angle is 90°.
Look at the bottom side of the rectangle. It intersects with two lines to form 3 angles: a 70° angle and two angles θ₁
Take the sum of the three angles and set the sum equal to 180°, then solve for θ₁:
2θ₁ + 70° = 180°
2θ₁ = 110°
θ₁ = 55°
We have two right triangles inscribed in the rectangle. Let's focus on the triangle on the right side. It consists of the angles θ₁, θ₃, and a right angle. Take the sum of the three angles and set the sum equal to 180° then solve for θ₃:
θ₁ + θ₃ + 90° = 180°
55° + θ₃ + 90° = 180°
θ₃ = 35°
The triangle on the left side is just the mirror image of the triangle on the right side. These triangles have the same angle measurements, so we can say right away that θ₄ = 35°
Angles θ₂ & θ₃ and θ₄ & θ₅ make up right angles, which have a total angle measurement of 90°. θ₃ = θ₄ = 35°, therefore we can say that θ₂ = θ₅ = 90° - 35°
θ₂ = θ₅ = 55°
B. There Is 4 Valence Electrons!
