Calculating PE and Initial Speed for Colliding Alpha Particles

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To calculate the minimum energy required for two colliding alpha particles, each must have an energy of 5.33 x 10^-46 eV. The mass of an alpha particle is approximately 6.64 x 10^-27 kg, and using the work-energy theorem, the initial speed needed for each particle to collide is 1.30 x 10^7 m/s. The potential energy when the particles are touching is derived from the formula PE = 1/2mv^2. This analysis provides the necessary calculations for understanding the collision dynamics of alpha particles.
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I need help for the following problem:

Two alpha-particles (diameter 1.9 x 10^-19 m) are headed directly toward each other with equal speeds. Compute the minimum energy in electron volts each particle must have if they are to collide. What inital speed must each particle have?

For the initial speed, i know that i need the work energy theorem. However, i don't know how to get the work. Thanks.
 
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Find the potential energy when the particles are touching each other.
Then find what their speeds would be far away from each other if they were released from this condition.
 


To calculate the minimum energy in electron volts (eV) for each alpha particle, we can use the equation PE = 1/2mv^2, where PE is the potential energy, m is the mass of the alpha particle, and v is the initial speed.

Since the two alpha particles have equal speeds, we can use the same formula for both particles. The mass of an alpha particle is approximately 6.64 x 10^-27 kg. Plugging in this value and the given diameter (1.9 x 10^-19 m) into the formula, we get:

PE = 1/2(6.64 x 10^-27 kg)(v^2)

To find the minimum energy in eV, we need to convert the units of mass and velocity into eV. We can use the conversion 1 eV = 1.602 x 10^-19 J. Therefore, the equation becomes:

PE = 1/2(6.64 x 10^-27 kg)(v^2)(1.602 x 10^-19 J/eV)

Simplifying, we get:

PE = 5.33 x 10^-46 v^2 eV

To solve for the initial speed, we can use the work-energy theorem, which states that the change in kinetic energy (KE) is equal to the work done on an object. In this case, the work done is the change in potential energy (PE) since the alpha particles are initially at rest.

Therefore, we can set the initial PE to be equal to the final KE:

PE = KE

Substituting the formula for PE that we found earlier, we get:

5.33 x 10^-46 v^2 eV = 1/2mv^2

Solving for v, we get:

v = √(5.33 x 10^-46 eV/m)

Converting the units back to m/s, we get:

v = 1.30 x 10^7 m/s

Therefore, each alpha particle must have a minimum energy of 5.33 x 10^-46 eV and an initial speed of 1.30 x 10^7 m/s in order to collide. I hope this helps with your problem!
 
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