A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 protons and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay?A : Positively charged. B : No particle is produced. C : Negatively charged. D : Neutral.

Positively charged particle

Explanation:

¹⁰₆C → ¹⁰₅B + e⁺ + ve

Since there is a decrease in the atomic number, there has been nuclear transmutation and a proton has be broken into a neutron and positron with the release of neutrino which has zero charge and very small mass. This is a beta plus decay with the release of Positively charged particle

(a) positively charged

Let's take the scenario step by step

(i) Carbon-10 with 6 protons and 4 neutrons. i.e

¹⁰₆C

(ii) Undergoes a radioactive beta decay (β) decay to form Boron-10 with 5 protons and 5 neutrons ¹⁰₅B as follows;

¹⁰₆C  =>  ¹⁰₅B + β   (i)

Remember that;

β decay can either be β- or β+

Where;

β -   =   ⁰₋₁ e⁻ (called electron and is negatively charged)

β +  =   ⁰₁ e⁺ (called positron and is positively charged)

Now, to balance the equation (i) above, β+ has to be used as follows;

¹⁰₆C =>  ¹⁰₅B + ⁰₁ e⁺

Therefore, the additional particle produced during this decay is ⁰₁ e⁺ (called positron and is positively charged)

c) up

hope this

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach