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Chapter 45: Problem 18
How many protons, neutrons, and electrons does an atom of \({ }_{92}^{235} \mathrm{U}\) possess?
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How many protons, neutrons, and electrons are there in \((a){ }^{3} \mathrm{He},(b){ }^{12} \mathrm{C}\), and \((c)^{206} \mathrm{~Pb}\) ? (a) The atomic number of He is 2 ; therefore, the nucleus must contain 2 protons. Since the mass number of this isotope is 3 , the sum of the protons and neutrons in the nucleus must equal 3 ; therefore, there is 1 neutron. The number of electrons in the atom is the same as the atomic number, \(2 .\) (b) The atomic number of carbon is 6 ; hence, the nucleus must contain 6 protons. The number of neutrons in the nucleus is equal to \(12-6=6 .\) The number of electrons is the same as the atomic number, \(6 .\) (c) The atomic number of lead is 82 ; hence, there are 82 protons in the nucleus and 82 electrons in the atom. The number of neutrons is \(206-82=124\).
The half-life of carbon- 14 is \(5.7 \times 10^{3}\) years. What fraction of a sample of \({ }^{14} \mathrm{C}\) will remain unchanged after a period of five half-lives?
By how much does the mass of a heavy nucleus change when it emits a 4.8-MeV gamma ray?
Complete the following nuclear equations: (a) \({ }^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{8}^{17} \mathrm{O}+\) ? (d) \({ }_{15}^{30} \mathrm{P} \rightarrow{ }_{14}^{30} \mathrm{Si}+\) ? (b) \({ }_{4}^{9} \mathrm{Be}+{ }_{2}^{3} \mathrm{He} \rightarrow{ }_{6}^{12} \mathrm{C}+?\) (e) \({ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+?\) (c) \({ }_{4}^{9} \operatorname{Be}(p, \alpha)\) ? (f) \({ }_{20}^{43} \mathrm{Ca}(\alpha, ?) \frac{46}{21} \mathrm{Sc}\) (a) The sum of the subscripts on the left is \(7+2=9 .\) The subscript of the first product on the right is 8 . Hence, the second product on the right must have a subscript (net charge) of \(1 .\) Also, the sum of the superscripts on the left is \(14+4=18 .\) The superscript of the first product is \(17 .\) Hence, the second product on the right must have a superscript (mass number) of 1 . The particle with nuclear charge 1 and mass number 1 is the proton, \({ }_{1}^{1} \underline{H}\). (b) The nuclear charge of the second product particle (its subscript) is \((4+2)-6=0 .\) The mass number of the particle (its superscript) is \((9+4)-12=1 .\) Hence, the particle must be the neutron, \(\downarrow{n}\). (c) The reactants \({ }_{4}^{9} \mathrm{Be}\) and \({ }_{1}^{1} \mathrm{H}\) have a combined nuclear charge of 5 and a mass number of \(10 .\) In addition to the alpha particle, a product will be formed of charge \(5-2=3\) and mass number \(10-4=6\). This is \(\frac{6}{3} \mathrm{Li}\). (d) The nuclear charge of the second product particle is \(15-14=+1\). Its mass number is \(30-30=0\). Hence, the particle must be a positron, \({ }_{+1} e\). (e) The nuclear charge of the second product particle is \(1-2=-1 .\) Its mass number is \(3-3=0 .\) Hence. the particle must be a beta particle (an electron), \(_{-1}^{0} e\). ( \(f\) ) The reactants, \(_{20}^{43} \mathrm{Ca}\) and \({ }_{2}^{4} \mathrm{He}\), have a combined nuclear charge of 22 and mass number of \(47 .\) The ejected product will have charge \(22-21=1\), and mass number \(47-46=1 .\) This is a proton and should be represented in the parentheses by \(p\). In some of these reactions a neutrino and/or a photon are emitted. We ignore them for this discussion since the charge for both are zero. Moreover the mass of the photon is zero and the mass of each of the several neutrinos, although not zero, is negligibly small.
How much energy must a bombarding proton possess to cause the reaction \({ }^{7} \operatorname{Li}(p, n)^{7} \mathrm{Be}\) ? Give your answer to three significant figures. The reaction is as follows: $$ { }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow{ }_{4}^{7} \mathrm{Be}+{ }_{0}^{1} n $$ where the symbols represent the nuclei of the atoms indicated. Because the masses listed in Table \(45-2\) include the masses of the atomic electrons, the appropriate number of electron masses \(\left(m_{e}\right)\) must be subtracted from the values given. Subtracting the total reactant mass from the total product mass gives the increase in mass as \(0.00176 \mathrm{u}\). (Notice that the electron masses cancel out. This happens frequently, but not always.) To create this mass in the reaction, energy must have been supplied to the reactants. The energy corresponding to \(0.00176 \mathrm{u}\) is \((931 \times 0.00176) \mathrm{MeV}=1.65 \mathrm{MeV}\). This energy is supplied as \(\mathrm{KE}\) of the bombarding proton. The incident proton must have more than this energy because the system must possess some KE even after the reaction, so that momentum is conserved. With momentum conservation taken into account, the minimum KE that the incident particle must have can be found with the formula $$ \left(1+\frac{m}{M}\right)(1.65) \mathrm{MeV} $$ where \(M\) is the mass of the target particle, and \(m\) that of the incident particle. Therefore, the incident particle must have an energy of at least $$ \left(1+\frac{1}{7}\right)(1.65) \mathrm{MeV}=1.89 \mathrm{MeV} $$
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