(1) 1H1 + 1H1 ---->1H2 + 1e0 + (2) 1H1 + 1H2 ---->2He3 + (3) 2He3 + 2He3 ---->2H
ID: 2058774 • Letter: #
Question
(1) 1H1 + 1H1 ---->1H2 + 1e0 +(2) 1H1 + 1H2 ---->2He3 +
(3) 2He3 + 2He3 ---->2He4 + 1H1 + 1H1
In these reactions 1e0 is a positron (mass = 0.000549 u), is a neutrino (mass approximatly 0 u), and is a gamma ray photon (mass = 0 u). Note that reaction (3) uses two helium nuclei, which are formed by two reactions of type (1) and two reactions of type (2). Calculate the total amount of energy generated by the proton-proton cycle. The atomic masses are 1H1 (1.007825 u), 1H2 (2.014102 u), 2He3 (3.016030 u), 2He4 (4.002603 u). Be sure to account for the fact that there are two electrons in two hydrogen atoms, whereas there is one electron in a single deuterium atom. The mass of one electron is 0.000549 u.
Explanation / Answer
In the given question u need not require to bother about electrons as the masses given are atomic masses and not nuclear masses. In the later case i.e. if given nuclear masses you must consider electrons mass separately. Loss of mass in reaction (1) Mass of reactants - mass of products = m(1H2) + m(1e0) - 2*m(1H1) = 0.000999 u Loss of mass in reaction (2) Mass of R - Mass of P = m(1H1) + m(1H2) - m(2He3) = 0.005897 u Loss of mass in (3) = 0.013807 u For 1 complete proton cycle, Loss in mass = 2*(1) + 2*(2) + (3) = 0.027599 u Energy = u * 931 MeV = 0.027599 * 931 MeV = 25.7 MeV = 4.11*1^-12 J
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