Question

Beryllium-8 is an unstable isotope and decays into two α particles, which are helium nuclei with mass 6.68×10−27kg . This decay process releases 1.5×10−14J of energy. For this problem, let's assume that the mass of the Beryllium-8 nucleus is just twice the mass of an α particle and that all the energy released in the decay becomes kinetic energy of the α particles.If a Beryllium-8 nucleus is at rest when it decays, what is the speed of the α particles after they are released?

Answer 1

Answer:

1498502.2 m/s

Explanation:

mass of the helium =  6.68×10−27kg

energy released during the decay =  1.5×10−14J

since all the energy released is converted to kinetic energy of the alpha particle then particle will have kinetic energy = 0.5 mh v²

1.5×10−14J = 2 ( 0.5 mh v²) where mh is mass of helium

1.5×10−14J  / ( 6.68×10−27kg) = v²

v = 1498502.2 m/s

Answer 2

Answer:

v = 1498502.2 m/s

Explanation:

All energy that was released is been converted to kinetic energy of the alpha particle, particle having a kinetic energy of = 0.5 mh v²

We are given the following as;

mass of the helium =  6.68×10−27kg

decay process release energy of =  1.5×10−14J

Calculating the speed of alpha particle after the release in this equation, we have

1.5×10−14J = 2 ( 0.5 mh v²)

Substituting the value of mh in the equation, we have

v² = 1.5×10−14J  / ( 6.68×10−27kg)

To eliminate the square root, we introduce square root to both sides

√v² = √1.5×10−14J  / ( 6.68×10−27kg)

v= 1498502.2 m/s