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)

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

its c i just checked it