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# Coulomb's Law problem watch

1. An alpha particle is emitted from a radioactive source with K.E of 4.8 MeV.

The alpha particle travels in a vacuum directly towards a gold (197/79 Au) nucleus, as illustrated in Fig. 5.1.

path of alpha particle ----->----------------------------O (gold nucleus)

Fig. 5.1.

For the closest approach of the alpha particle to the gold nucleus determine

1. their separation.

I did exactly the steps from mark scheme except where they multiplied the charge-squared by 2,

P.E = Q1Q2/4pi(permittivity)r

and then equate that to 4.8 MeV. to get r value. But In the mark scheme, it stated P.E = (79x2(Q1Q2))/(4pi(permittivity)r)

Where does the '2' come from, I know why 79 because there are 79 protons but why into '2'? Please help. Thanks
2. somebody??
3. Okay, so you have the equation:

The energy will be 4.8MeV, but you have to convert it to Joules first:

( is the magnitude of the charge on the electron).

is the charge on the particle, and is the charge on the nucleus, so:

Then solve for ...

Does that not give the right answer?
4. (Original post by Taffss)

Where does the '2' come from, I know why 79 because there are 79 protons but why into '2'? Please help. Thanks
There are 2 positive charges on the alpha.
5. (Original post by Qwertish)
Okay, so you have the equation:

The energy will be 4.8MeV, but you have to convert it to Joules first:

( is the magnitude of the charge on the electron).

is the charge on the particle, and is the charge on the nucleus, so:

Then solve for ...

Does that not give the right answer?
Thannk youuu! =DD except I am not much familiar with eV to Joules maybe because I haven't covered charged particles chapter on my A2 book. But thanks a lot =D
6. (Original post by Stonebridge)
There are 2 positive charges on the alpha.
clear! =D

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