Use the following information to answer this analytic question. A plutonium-239

Question

Use the following information to answer this analytic question. A plutonium-239 nucleus, initially at rest, undergoes alpha decay to produce a uranium-235 nucleus. The uranium-235 nucleus has a mass of 390 x10-25 kg, and moves away from the location of the decay with a speed of 2.62 x 105 m/s. Determine the minimum electric potential difference that is required to bring the alpha particle to rest. Marks will be awarded based on the two physics principles (listed below) you choose, the formulas you state, the substitutions you show, and your final answer Uniform motion (balanced forces) Uniformly accelerated motion (unbalanced forces) Circular motion (unbalanced forces) Work-energy theorem .Conservation of momentum .Conservation of energy Conservation of mass-energy Conservation of charge .Conservation of nucleons Wave-particle duality

Explanation / Answer

we know,

m_alfa = 4*1.67*10^-27 kg

Apply conservation of momentum

momentum gained by alfa particle = momentum giened by Uranium atom

m_alfa*v = 3.90*10^-25*2.62*10^5

==> v = 3.90*10^-25*2.62*10^5/(4*1.67*10^-27)

= 1.53*10^7 m/s

let V is the required potential difference to stop the alfa particle.

now use, Work-Energy therem

Workone on alfa particle = loss of kinetic energy

q*V = (1/2)*m_alfa*v^2

V = (1/2)*m_alfa*v^2/q

= (1/2)*4*1.67*10^-27*(1.53*10^7)^2/(1.6*10^-19)

= 4.90*10^6 V <<<<<------Answer

here we neet to use Work-energy theorem and conservation of momentum

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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