Minimum distance & electrostatic potential energy

[victoriafello]

An alpha particle (made of 2 protons and 2 neutrons) with initial kinetic energy of 1.20 × 10^-12 J travels directly towards the nucleus of the stationary zinc atom The nucleus of a zinc atom contains 30 protons and 35 neutrons, and has a diameter of 4.00 × 10^-13 cm.

The alpha particle slows as kinetic energy is converted to electrostatic potential energy. Calculate the minimum distance between the centre of the zinc nucleus and the alpha particle. For simplicity, assume the charge of the nucleus to be concentrated at its centre. Express your answer in terms of the radius of the zinc nucleus

i can't put down my thoughts so far as i don't know where to start, i can't find an equation that let's you find distance with the data given so far, i would need to know the charge on the zinc & alpha particles ? i guess i can work this out from the charges on the protons & neutrons ?

 

[ehild]

The alpha particle is made of two protons and two neutrons. What is the charge of a proton? And what is the charge of a neutron? Can you find out what is the charge of the alpha particle then?
The same for the zinc nucleus. It contains 30 protons and 35 neutrons. What is its charge?

ehild

 

[victoriafello]

ok so

charge on a proton = 1.602x10^-19 C
charge on a neutron = 0

so total charge on the zinc is
30 x 1.602x20^-19 C = 4.806x10^-19

for the alpha particle
2 x 1.602x10^-19 C = 3.2x10^-19 C

now i need an equation for distance, this equation looks the closest ?

Eel = q1q2/4.Pi.Eo.r^2

but i have 2 unknowns Eel & r, unless i can get Eel from the kinetic energy

 

[ehild]

"so total charge on the zinc is
30 x 1.602x20^-19 C = 4.806x10^-19" wrong, it is 4.806x10^-18 C.

As the problem does not state the position of the alpha particle where its kinetic energy is given you can assume that it is very far away, where its potential energy is zero. The particle moves towards the nucleus and its potential energy increases, up to the point, where all its initial kinetic energy is transformed to potential energy. You have to find the distance from the nucleus where this happens.

ehild

 

[coz]

"

As the problem does not state the position of the alpha particle where its kinetic energy is given you can assume that it is very far away, where its potential energy is zero. The particle moves towards the nucleus and its potential energy increases, up to the point, where all its initial kinetic energy is transformed to potential energy. You have to find the distance from the nucleus where this happens.

ehild

Does this mean you can assume E_el is zero and rearrange the equation
E_el = q_1xq_2/4piE_or
so that r is the subject and work out the answer that way?
Thanks

 

[coz]

Does this mean you can assume E_el is zero and rearrange the equation
E_el = q_1xq_2/4piE_or
so that r is the subject and work out the answer that way?
Thanks

I don't think my above thoughts would work, as E_el would only equal zero when r = infinity and I am trying to work out r (in terms of the radius of the zinc nucleus). Back to the drawing board! :)

 

[ehild]

What do you mean on E_el?

The energy of the alpha particle is the sum of its kinetic energy and electric potential energy. This sum is equal to its initial kinetic energy. When its stops the KE =0, and the entire energy is electric. Find r where this happens.

ehild

 

[coz]

Sorry, I meant electrostatic potential energy when I said E_el

 

[coz]

What do you mean on E_el?

The energy of the alpha particle is the sum of its kinetic energy and electric potential energy. This sum is equal to its initial kinetic energy. When its stops the KE =0, and the entire energy is electric. Find r where this happens.

ehild

The initial kinetic energy is 1.20x10^-12 J, so the electric energy is also 1.20x10^-12 J.

Can I rearrange E=q₁q₂/4$$\pi$$$$\epsilon$$₀r to find r?

 

[ehild]

So find r where

$$E_{el}=\frac{1}{4\pi\epsilon_0}\frac{q1*q2}{r}=1.2*10^{-12 }J$$

Yes, you can rearrange to get r.

ehild

 

[coz]

So find r where

$$E_{el}=\frac{1}{4\pi\epsilon_0}\frac{q1*q2}{r}=1.2*10^{-12 }J$$

Yes, you can rearrange to get r.

ehild

Cool, thank you so much. I get r = 1.15x10^-14 m

Just one more thing, the question says to express your answer in terms of the radius of the zinc nucleus. I'm not sure what this means. The radius of the zinc nucleus is 2.00x10^-14 m.

Thanks again :)

 

[ehild]

It means r/R(Zn)= 0.575 R(Zn).

So the alpha particle goes into the Zn nucleus almost halfway to its centre. Of course, it is not true really, because the situation is completely different inside the nucleus and outside it. That poor zinc nucleus might explode, I am afraid.

 

[Kawakaze]

Just a thought, isn't 2.00x10^-13cm actually 2.00x10^-15 metres?

 

[ehild]

Just a thought, isn't 2.00x10^-13cm actually 2.00x10^-15 metres?

Ooops... you are right. I did not look at the first post. So the nucleus will survive.

ehild

 

[Kawakaze]

I hope you are wrong there. I just had a paper with this question on and I already submitted it!

r/R(Zn) = $$\frac{1.15 x 10^-14}{2.0 x 10^-15}$$

Which i got to be 5.75. or enough to reach the centre of the nucleus.

Or is it R(Zn)/r, I have to admit i don't really get this part of the question.

 

[jacobkeif]

Help... please explain how a nucleus containing 65 ions be smaller in diameter than one containing 4? Am i just confused here.