A beryllium-8 atom at rest undergoes double alpha decay as follows: 8 4 Be ? 4 2

Question

A beryllium-8 atom at rest undergoes double alpha decay as follows:
8
4
Be ?
4
2
He +
4
2
He
The atomic masses are:

He 4.002603

Be 8.005305
The kinetic energy of each departing alpha particle, in keV, is closest to:
A) 92 B) 180 C) 65 D) 46 E) 130

I know the answer is 46 I just need to know why and the work for it.

Explanation / Answer

Be----------> 2 He is eq Binding energy = (8.005305-2(4.002603)) x 931.5 MeV = 0.0922 MeV = 92.2 KeV BE= KE of two alfa particles 92.2 = 2 KE of alfa particle KE of alfa particle = 46 kev hence D is ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.