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Chapter 29: Problem 52

The reactions listed in Problem 51 did not stop there. To the surprise of the Curies, the phosphorus decay continued after the \(\alpha\) bombardment ended with the phosphorus \(^{30}\) is \(\mathrm{P}\) emitting a \(\beta^{+}\) to form yet another product. Write out this reaction, identifying the other product.

Short Answer

Expert verified
Answer: The nuclear reaction when phosphorus-30 emits a positron is: \(_{15}^{30}\mathrm{P}\) -> \(_{14}^{30}\mathrm{Si}\) + \(_{1}^{0}\beta^{+}\)

Step by step solution

01

Write Down the General Nuclear Reaction Form

We will start by writing down the general form of a nuclear reaction, which is given as follows: _PRODUCTION_ + _PARENT_ -> _DAUGHTER_ + _EMISSION_ In this exercise, our PARENT is the phosphorus-30 (\(^{30}\mathrm{P}\)) and the EMISSION is the positron (\(\beta^{+}\)).
02

Determine the Number of Nucleons in Phosphorus-30 and the Positron

To find the number of nucleons in \(^{30}\mathrm{P}\), we simply look at the mass number which is 30. For the positron, we need information about its charge and mass. A positron is an antiparticle of the electron, and it has the same mass as an electron but with a positive charge. Electrons and positrons have negligible mass compared to protons and neutrons, so it effectively has 0 nucleons.
03

Write Down the Nuclear Reaction with Nucleons Conserved

Now, we will write down the nuclear reaction while conserving the nucleons we determined in Step 2: _PRODUCTION_ + \(_{15}^{30}\mathrm{P}\) -> _DAUGHTER_ + \(_{1}^{0}\beta^{+}\) _Phosphorus-30_ -> _DAUGHTER_ + _Positron_
04

Determine the Other Product (DAUGHTER) of the Reaction

To find the other product (DAUGHTER) of the reaction, we simply conserve nucleons. Since neutral phosphorus has 15 protons and 15 neutrons, emitting a positron means that it loses a positive charge. Losing a positive charge is equivalent to losing a proton which is shown as follows: \(_{15}^{30}\mathrm{P}\) -> _DAUGHTER_ + \(_{1}^{0}\beta^{+}\) So, the phosphorus-30 loses one proton and becomes an element with one less proton. The element with an atomic number of 14 is silicon (Si). The mass number remains the same since nucleons are conserved, so the daughter nucleus is silicon-30: \(_{15}^{30}\mathrm{P}\) -> \(_{14}^{30}\mathrm{Si}\) + \(_{1}^{0}\beta^{+}\)
05

Write Down the Final Nuclear Reaction

Now that we have the other product (DAUGHTER) of the reaction, we can write down the final nuclear reaction: \(_{15}^{30}\mathrm{P}\) -> \(_{14}^{30}\mathrm{Si}\) + \(_{1}^{0}\beta^{+}\) So, when phosphorus-30 decays by emitting a positron, it forms silicon-30.

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Most popular questions from this chapter

Suppose that a radioactive sample contains equal numbers of two radioactive nuclides \(A\) and \(B\) at \(t=0 .\) A has a half-life of \(3.0 \mathrm{h},\) while \(\mathrm{B}\) has a half-life of \(12.0 \mathrm{h}\) Find the ratio of the decay rates or activities \(R_{\mathrm{A}} / R_{\mathrm{B}}\) at (a) \(t=0,\) (b) \(t=12.0 \mathrm{h},\) and \((\mathrm{c}) t=24.0 \mathrm{h}\).
An isotope of sodium, \({ }_{11}^{2} \mathrm{Na}\), decays by \(\beta^{+}\) emission. Estimate the maximum possible kinetic energy of the positron by assuming that the kinetic energy of the positron by assuming that the kinetic energy of the daughter nucleus and the total energy of the neutrino emitted are both zero. [Hint: Remember to keep track of the electron masses.
Identify the daughter nuclide when \(_{19}^{40} \mathrm{K}\) decays via \(\beta^{-}\) decay.
Write out the reaction and identify the daughter nuclide when \(_{11}^{22}\) Na decays by electron capture.
Write out the reaction and identify the daughter nuclide when \(_{11}^{22}\) Na decays by emitting a positron.
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