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Chapter 19: Problem 18

Write balanced equations for each of the processes described below. a. Chromium-51, which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- \(32,\) which accumulates in the liver, decays by \(\beta\) -particle production.

Short Answer

Expert verified
a. \( _{24}^{51}Cr + _{-1}^0e \rightarrow _{23}^{51}Mn \) b. \( _{53}^{131}I \rightarrow _{54}^{131}Xe + _{-1}^0e \) c. \( _{15}^{32}P \rightarrow _{16}^{32}S + _{-1}^0e \)

Step by step solution

01

Process a: Chromium-51 decays by electron capture

1. The initial element is chromium-51 (Cr), which has an atomic number Z = 24 and a mass number A = 51. Since it decays by electron capture, it will transform into an element with an atomic number of 23. 2. In electron capture, an inner orbital electron is captured by the nucleus, causing the atomic number to decrease by 1 (Z = 23) and the mass number remains the same (A = 51). 3. The equation is: \( _{24}^{51}Cr + _{-1}^0e \rightarrow _{23}^{51}X \) The final element is manganese (Mn), which has an atomic number of 23, so the balanced nuclear equation becomes: \( _{24}^{51}Cr + _{-1}^0e \rightarrow _{23}^{51}Mn \)
02

Process b: Iodine-131 decays by producing a beta particle

1. The initial element is iodine-131 (I), which has an atomic number Z = 53 and a mass number A = 131. Since it decays by producing a beta particle (electron), it will transform into an element with an atomic number of 54. 2. In beta decay, a neutron in the nucleus is converted to a proton and an electron (beta particle). The atomic number increases by 1 (Z = 54), and the mass number remains the same (A = 131). 3. The equation is: \( _{53}^{131}I \rightarrow _{54}^{131}X + _{-1}^0e \) The final element is xenon (Xe), which has an atomic number of 54, so the balanced nuclear equation becomes: \( _{53}^{131}I \rightarrow _{54}^{131}Xe + _{-1}^0e \)
03

Process c: Phosphorus-32 decays by beta-particle production

1. The initial element is phosphorus-32 (P), which has an atomic number Z = 15 and a mass number A = 32. Since it decays by beta-particle production, it will transform into an element with an atomic number of 16. 2. Again, as a beta decay, a neutron in the nucleus is converted to a proton and an electron (beta particle). The atomic number increases by 1 (Z = 16), and the mass number remains the same (A = 32). 3. The equation is: \( _{15}^{32}P \rightarrow _{16}^{32}X + _{-1}^0e \) The final element is sulfur (S), which has an atomic number of 16, so the balanced nuclear equation becomes: \( _{15}^{32}P \rightarrow _{16}^{32}S + _{-1}^0e \)

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Most popular questions from this chapter

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