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balu736 [363]
4 years ago
5

A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 proto

ns 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.
Physics
2 answers:
Studentka2010 [4]4 years ago
6 0
<h2>Answer:</h2>

(a) positively charged

<h2>Explanation:</h2>

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)

<em>Remember that;</em>

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)

svetoff [14.1K]4 years ago
5 0

Answer:

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

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Read 2 more answers
The temperature of air changes from 0 to 18°C while its velocity changes from zero to a final velocity, and its elevation change
irina [24]

Answer:

For the air:

Final Velocity 160.77m/s

Final Elevation 1,317.43m

the Internal, Kinetic, and Potential Energy changes  will be equal.

Explanation:

In principle we know the following:

  • <u>Internal Energy:</u> is defined as the energy contained within a system (in terms of thermodynamics). It only accounts for any energy changes due to the internal system (thus any outside forces/changes are not accounted for). In S.I. is defined as U=mC_{V}\Delta T where m is the mass (kg), C_{V} is a specific constant-volume (kJ/kg°C) and \Delta T is the Temperature change in °C.
  • <u>Kinetic Energy:</u> denotes the work done on an object (of given mass m) so that the object at rest, can accelerate to reach a final velocity. In S.I. is defined as K=\frac{1}{2}mv^2 where v is the velocity of the object in (m/s).
  • <u>Potential Energy:</u> denotes the energy occupied by an object (of given mass m) due to its position with respect to another object. In S.I. is defined as P=mgh, where g is the gravity constant equal to 9,81m/s^2 and h is the elevation (meters).

<em>Note: The Internal energy is unaffected by the Kinetic and Potential Energies.</em>

<u>Given Information:</u>

  • Temperature Change 0°C → 18°C ( thus \Delta T=18°C )
  • Object velocity we shall call it v_{o} and v_{f}, for initial and final, respectively. Here we also know that v_{o}=0m/s^2
  • Object elevation we shall call it h_{o} and h_{f}, for initial and final, respectively. Here we also know that h_{o}= 0m

∴<em> We are trying to find v_{f} and h_{f} of the air where U, K and P are equal.</em>

Lets look at the change in Energy for each.

<u>Step 1: Change in Kinetic Energy=Change in Internal Energy</u>

\Delta E_{K}=\Delta U\\\frac{1}{2}m{v_{f}}^2- \frac{1}{2}m{v_{o}}^2=mC_{V}\Delta T

Here we recall that v_{o}=0m/s^2 and mass m is the same everywhere. Thus we have:

\frac{1}{2}m{v_{f}}^2=mC_{V}\Delta T    

\frac{1}{2} {v_{f}}^2=C_{V}\Delta T\\ {v_{f}}^2=2C_{V}\Delta T\\ v_{f}=\sqrt{2C_{V}\Delta T}     Eqn(1)

<u>Step 2: Change in Potential Energy=Change in Internal Energy</u>

\Delta E_{P}=\Delta U\\mgh_{f}-mgh_{o}=mC_{V}\Delta T

Here we recall that h_{o}=0m/s^2 and mass m is the same everywhere. Thus we have:

mg(h_{f}-h_{o})=mC_{V}\Delta T\\gh_{f}=C_{V}\Delta T\\

h_{f}=\frac{C_{V}\Delta T}{g}      Eqn(2).

Finally by plugging the known values in Eqns (1) and (2) we obtain:

v_{f}=\sqrt{2*718*18}=160.77m/s

h_{f}=\frac{718*18}{9.81}=1,317.43m

Thus we can conclude that for the air final velocity v_{f}=160.77m/s and final elevation h_{f}=1,317.43m the internal, kinetic, and potential energy changes  will be equal.

3 0
3 years ago
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