Alpha Particle Approaching a Gold Nucleus

In summary, the question asks for the distance of the closest approach between an alpha-particle and a gold nucleus, assuming Coulomb's law and conservation of momentum and energy. The calculation involves finding the velocity of the gold nucleus and using it to equate the change in kinetic energy to the change in electric potential energy. The gold nucleus is assumed to be a fixed field/potential generator while the alpha-particle is in motion.
  • #1
FelaKuti
19
0

Homework Statement


An alpha-particle with velocity 3.5 x 106 strikes a block of gold, atomic number 79 and mass number 197. Find the distance of the closest possible approach between the alpha particle and a gold nucelus, assuming Coulomb's law holds over such distances. Assume mp = mn = 1.67 x 10-27

Homework Equations


p = mv
KE = 1/2mv2
EPE = kq1q2/r2

The Attempt at a Solution



Considering conversation of momentum and conservation of energy I worked out p = mv = (4 x 1.67 x 10-27)(3.5 x 106) = 2.3 x 10-20.

The particle would repel the nucleus, at the closest approach they must have the same velocity. So p = mauv' + malpha v'
v' = p/mau + mualpha
v' = 2.3 x 10-20 / 201 x 1.67 x 10-27
v' = 6.4 x 104

So the change in KE in KEv minus KEv'.

At this point I believe I've gone wrong as I've calculated the change in KE to be negligible yet the online question pp I'm using tells me to consider conversation of momentum (here). As I understand the remainder of the question is simply equate change KE to change in EPE, rearrange to find r and plug in values.
 
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  • #2
The gold nucleus is to be assumed as a field/potential generator and hold firm in its initial place. In this case, only conservation of energy is needed. Start from infinite distance where the alpha particle is moving with the given velocity in zero potential. Then the total energy in the closest distance will give you the answer.
 
  • #3
blue_leaf77 said:
The gold nucleus is to be assumed as a field/potential generator and hold firm in its initial place.
I don't believe that's true in this instance. If you check the question and hints at the website you'll find that the use of conservation of momentum is recommended in addition to conservation of energy.

@FelaKuti , check your calculation for v'. Your method's okay but the numerical result looks off; maybe a finger problem on the calculator.
 
  • #4
I see. I only read the first hint where no mention of momentum conservation is made. So yes the momentum conservation is needed and the gold nucleus is also moving.
 

FAQ: Alpha Particle Approaching a Gold Nucleus

1. What is an alpha particle?

An alpha particle is a type of nuclear radiation that consists of two protons and two neutrons, making it identical to a helium nucleus.

2. How does an alpha particle approach a gold nucleus?

An alpha particle approaches a gold nucleus through a process called alpha decay, where the unstable nucleus of an atom emits an alpha particle to become more stable.

3. What happens when an alpha particle approaches a gold nucleus?

When an alpha particle approaches a gold nucleus, they may either collide and fuse together, or they may be deflected by the strong positive charge of the gold nucleus due to electromagnetic repulsion.

4. What is the significance of studying alpha particles approaching gold nuclei?

Studying alpha particles approaching gold nuclei can help us understand the fundamental properties of atoms and how they interact with each other. This research can also have practical applications in fields such as nuclear energy and medicine.

5. Can alpha particles approaching gold nuclei have any harmful effects?

Alpha particles approaching gold nuclei can be harmful if they are emitted from a radioactive source and come into contact with living cells, causing damage to their DNA. However, in controlled research settings, the levels of radiation are kept low and do not pose a significant risk to scientists studying these particles.

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