Electric Potential Difference involving Alpha Decay question

In summary, the problem requires determining the minimum electric potential difference needed to bring an alpha particle to rest, which is calculated using the equation V = E/q. The initial mass of the plutonium-239 nucleus is subtracted from the mass of the products to get the mass of the alpha particle, which is then used to calculate the energy (E=mc^2) and charge (q=elementary charge x number of protons). The final equation is V = E/q, with a solution of 6.08 x 10^33 V. Other approaches, such as using the kinetic energy formula, may also be possible.
  • #1
Eloc Jcg
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0

Homework Statement


A plutonium-239 nucleus, initially at rest, undergoes alpha decay to produce a uranium-235 nucleus. The uranium-235 nucleus has a mass of 3.90x10^-25 kg, & moves away from the location of the decay with a speed of 2.62x10^5 m/s. Determine the minum electric potential difference that is required to bring the alpha particle to rest.


Homework Equations


V = ? = E/q
E = mc^2
m = mparent - mproducts
q = elementry charge x #protons

The Attempt at a Solution


m = 239.052156u - 235.0439299u - 4.002603u = -0.9943769u
E = -0.9943769u x 3x10^8m/s = 8.9493921x10^16 eV
q = 1.6x10^19C x 92protons = 1.472x10^-17 C

V = 8.9493921x10^16eV / 1.472x10^-17C = 6.08 x 10^33 V

Am I way off here in my approach to solving this problem? Or should this problem be solved using the Ek = 1/2mv^2 formula? Any help is appreciated as this is my last assignment I am needing to do, which assuming I pass my final, I will be done high school. I've been up all night working on this problem, & flipping through my high school physics textbook & my college electronics textbook, but I am not sure what to expect as my final answer.

Thanks.
 
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  • #2
I was thinking maybe use the F = E q = Vq/d = ma ?
But that means any non zero potential will be able to bring it to rest (lower potential will simply take long to do it and the nucleus will travel a larger distance while decelerating).
 

1. What is electric potential difference?

Electric potential difference is the change in electric potential energy per unit charge between two points in an electric field. It is measured in volts (V).

2. How is electric potential difference involved in alpha decay?

Alpha decay is a type of radioactive decay in which an atom emits an alpha particle, consisting of two protons and two neutrons. This process creates a difference in electric potential between the parent atom and the resulting daughter atom, as the alpha particle carries a positive charge.

3. Can electric potential difference be calculated for alpha decay?

Yes, electric potential difference can be calculated for alpha decay by using the equation V = kq/r, where V is the potential difference, k is Coulomb's constant, q is the charge of the alpha particle, and r is the distance between the parent and daughter atom.

4. How does electric potential difference affect the rate of alpha decay?

The higher the electric potential difference between the parent and daughter atom, the greater the force of repulsion between them. This can lead to a higher rate of alpha decay as the parent atom tries to reach a more stable state.

5. How is electric potential difference related to the half-life of a radioactive substance?

The half-life of a radioactive substance is the time it takes for half of the substance to decay. Electric potential difference can affect this process by influencing the rate of decay. A higher electric potential difference can lead to a shorter half-life, as there is a greater force driving the decay process.

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