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    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
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    somebody??
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    Okay, so you have the equation:

    

E = \dfrac{Q_1 Q_2}{4 \pi \varepsilon_0 r}

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

    

4.8MeV = 4.8e\times 10^{6} = 7.69\times 10^{-13}J

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

    Q_1 is the charge on the \alpha particle, and Q_2 is the charge on the nucleus, so:

    

4.8e\times 10^{6} = \dfrac{(2e)(79e)}{4 \pi \varepsilon_0 r}

    Then solve for r...

    Does that not give the right answer?
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    (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.
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    (Original post by Qwertish)
    Okay, so you have the equation:

    

E = \dfrac{Q_1 Q_2}{4 \pi \varepsilon_0 r}

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

    

4.8MeV = 4.8e\times 10^{6} = 7.69\times 10^{-13}J

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

    Q_1 is the charge on the \alpha particle, and Q_2 is the charge on the nucleus, so:

    

4.8e\times 10^{6} = \dfrac{(2e)(79e)}{4 \pi \varepsilon_0 r}

    Then solve for r...

    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
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    (Original post by Stonebridge)
    There are 2 positive charges on the alpha.
    clear! =D
 
 
 
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