The atomic mass of 14 C 14C is 14.003242 14.003242 u. Show that the β − \beta^{-} decay of 14 C 14C is energetically possible, and calculate the energy released in the decay.

Hi, everyone in this practice problem, we're being asked to show that the beta plus the K of 22 Nesum or N A according to the equation of 22 11 and a decaying into 22 10 neon and E plus zero plus one E or positron plus V E is energetically possible. And we're also being asked to calculate the energy produced. 22 N A has an atomic mass of 21.994438 U and 22 N E has an atomic mass of 21.9913851 U A positron has a mass of 0. U. And the options given for the energy produced are a 1.15 mega E V B 2.33 mega E V C 4.15 mega E V and D 7.65 mega E V. So we will start by comparing the total mass on each side of the decay equation. So the mass of the para nuclei for 22 N A or M 22 N A will be equal to 21. 38 U. And then the mass of the decay product or 22 N E and the Positron M 22 N E plus zero plus one, E will be equal to 21. U plus 0. U which will come up to a value of 21. U. So the, so the mass of the parent newly is larger than the mass of the delta nu, therefore, the decay is going to be energetically possible. So the energy released during the decay can be calculated by E equals to delta M multiplied by C squared where E can then be calculated by delta M or the difference in mass, which is going to be the mass of the parent nude 21.994438 U minus 21.9919337 U or the mass of the daughter nude plus Positron. So in this case, we can then uh multiply this with the C squared which will be 931.5 mega E V divided by U. And then calculating this, our E will come out to be uh 2.5 oh four times 10 to the power of negative three U multiplied by 931.5 mega E V per U and calculating this, the energy will come out to be 2.33 mega E V. So the energy released during the decay is going to be 2.33 M E V or mega E V which will correspond to option B in our answer choices. So option B will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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Problem 15

Textbook Question

The atomic mass of $14 C$ is $14.003242$ u. Show that the $β^{-}$ decay of $14 C$ is energetically possible, and calculate the energy released in the decay.

Verified step by step guidance

Determine the nuclear reaction for the beta-minus (β⁻) decay of 14C. In β⁻ decay, a neutron in the nucleus converts into a proton, emitting an electron (β⁻ particle) and an antineutrino. The reaction is:

C^{14} → N^{14} + β⁻ + ν̅.

Look up the atomic mass of the daughter nucleus, nitrogen-14 (14N). The atomic mass of 14N is 14.003074 u. Note that the atomic masses include the mass of electrons, so we do not need to separately account for the emitted β⁻ particle's mass.

Calculate the mass difference (Δm) between the parent nucleus (14C) and the daughter nucleus (14N). Use the formula:

Δm = m_{C 14} - m_{N 14}

. Substitute the given values:

= 14.003242 \, ext{u} - 14.003074 \, ext{u}

Convert the mass difference (Δm) into energy using Einstein's mass-energy equivalence formula:

E = Δm \, c^{2}

. Here,

c

is the speed of light. To convert atomic mass units (u) to energy, use the conversion factor:

1 \, 	ext{u} = 931.5 \, 	ext{MeV/c}^2

. Thus,

= Δm \, imes \, 931.5 \, ext{MeV}

Interpret the result: If the calculated energy (E) is positive, the β⁻ decay of 14C is energetically possible. The positive energy value represents the energy released during the decay process.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Beta Decay

Beta decay is a type of radioactive decay in which a beta particle (an electron or positron) is emitted from an atomic nucleus. In the case of carbon-14 (14C), it undergoes beta decay to transform into nitrogen-14 (14N) by converting a neutron into a proton, which increases the atomic number by one while keeping the mass number the same.

Mass-Energy Equivalence

Mass-energy equivalence, expressed by Einstein's equation E=mc², states that mass can be converted into energy and vice versa. In nuclear reactions, the difference in mass between the reactants and products can be converted into energy, which is released during processes like beta decay. This principle is crucial for calculating the energy released in the decay of 14C.

Q-Value of a Decay

The Q-value of a decay is the amount of energy released during a nuclear reaction, calculated as the difference in mass between the initial and final states, multiplied by c². For the beta decay of 14C, the Q-value can be determined by comparing the atomic masses of 14C and 14N, along with the emitted beta particle, allowing us to quantify the energy released in the process.

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The atomic mass of 14C14C is 14.00324214.003242 u. Show that the β−\beta^{-} decay of 14C14C is energetically possible, and calculate the energy released in the decay.

Key Concepts

Beta Decay

Mass-Energy Equivalence

Q-Value of a Decay

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The atomic mass of $14 C$ is $14.003242$ u. Show that the $β^{-}$ decay of $14 C$ is energetically possible, and calculate the energy released in the decay.