Question

In a location in outer space far from all other objects, a nucleus whose mass is 4.019480 × 10−25 kg and that is initially at rest undergoes spontaneous alpha decay. The original nucleus disappears, and two new particles appear: a He-4 nucleus of mass 6.640678 × 10−27 kg (an alpha particle consisting of two protons and two neutrons) and a new nucleus of mass 3.952926 × 10−25 kg. These new particles move far away from each other, because they repel each other electrically (both are positively charged). Because the calculations involve the small difference of (comparatively) large numbers, you need to keep seven significant figures in your calculations, and you need to use the more accurate value for the speed of light, 2.9979246e8 m/s. Choose all particles as the system. Initial state: Original nucleus, at rest. Final state: Alpha particle + new nucleus, far from each other.

Part 1 (a) What is the rest energy of the original nucleus? (Round your answer to seven significant figures.) J the tolerance is +/-2%

Part 2 (b) What is the sum of the rest energies of the alpha particle and the new nucleus? (Round your answer to seven significant figures.) J the tolerance is +/-2%

Part 3 (c) Did the portion of the total energy of the system contributed by rest energy increase or decrease? decrease The total rest energy is unchanged. increase

Part 4 (d) What is the sum of the kinetic energies of the alpha particle and the new nucleus?

Answer 1

Solution or Explanation:

speed of light, 2.9979246e8 m/s

He-4 nucleus of mass 6.640678 × 10−27 kg

initial nucleus mass is 4.019480 × 10−25 kg

New nucleus of mass 3.952926 × 10−25 kg.

Ei= m1c2

Ef= m2c2+ K2+ m3c2+ K3

Taking the system as everything, Wext= 0

Therefore Ef= Ei, and you can solve for K2+ K3= m1c2- m2c2- m3c2

(a) rest energy of the original nucleus = m₀c² = 4.019480 × 10−25 x (2.9979246e8)²

= 3.612527 x 10^-8 j

(b) sum of the rest energies of the alpha particle and the new nucleus

m₁c² = 6.640678 × 10−27 x (2.9979246e8)² =5.9683438 x 10^-10 j

m₂c² = 3.952926 × 10−25 x (2.9979246e8)² = 3.5527114 x 10^-8 j

sum = m₁c² + m₂c² = (5.9683438 x 10^-10 + 3.5527114 x 10^-8 )j = 3.6123948 x 10^-8 J

(c) The portion of the total energy of the system contributed by rest energy: decreased

Therefore the portion of the total energy of the system contributed by kinetic energy: increased

(d) sum of the kinetic energies of the alpha particle and the new nucleus

Kalpha+ Knew nucleus = 3.612527 x 10^-8 J -3.6123948 x 10^-8 J

K.E = 1.322 x 10^-12 J