Determine the kinetic energy of Be7 nuclei emerging

To determine the kinetic energy of Be7 nuclei arising in the reaction p+Li7 = Be7+n, Q= - 1.64 MeV at the threshold value of the kinetic energy of the bombarding particle (proton and neutron), it is necessary to solve problem 60334. Formulas and laws should be used in it , used to calculate the kinetic energy of nuclei. The result of the solution will be the output of the calculation formula and the answer. If you have any questions about the solution, please contact us - we will try to help.

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In this problem, it is necessary to determine the kinetic energy of Be7 nuclei that arise in the reaction p+Li7=Be7+n, where Q=-1.64 MeV, at a threshold value of the kinetic energy of the bombarding particle (proton and neutron).

To solve the problem it is necessary to use the law of conservation of energy. During the reaction, mass is converted into energy according to the formula E=mc^2, where E is energy, m is mass, c is the speed of light. It is also necessary to use the formula for the kinetic energy of the particle, which is expressed as E_kin=(mv^2)/2, where m is the mass of the particle, v is its speed.

Let's calculate the mass of the Be7 nucleus, which is equal to the sum of the masses of the proton and lithium, expressed in atomic mass units (u): m(Be7) = m(p) + m(Li7) = 1.007276 u + 7.016004 u = 8.02328 u.

It is also necessary to calculate the mass of the Be7+n nucleus, which is equal to the sum of the masses of the Be7 nucleus and the neutron: m(Be7+n) = m(Be7) + m(n) = 8.02328 u + 1.008665 u = 9.031945 u.

Since the reaction occurs at a threshold value of the kinetic energy of the bombarding particle, the reaction energy will be zero. Therefore, the reaction energy is equal to the difference between the masses of the initial and final nuclei, multiplied by the speed of light squared: E = (m(p) + m(Li7) - m(Be7) - m(n)) * c^2 = (1, 007276 u + 7.016004 u - 8.02328 u - 1.008665 u) * (2.99792*10^8 m/s)^2 = 4.417 MeV.

Since Q = -1.64 MeV, the kinetic energy of Be7 nuclei will be equal to the difference between the reaction energy and Q: E_kin(Be7) = E - Q = 4.417 MeV - (-1.64 MeV) = 6.057 MeV.

Answer: the kinetic energy of Be7 nuclei arising in the reaction p+Li7=Be7+n, at the threshold value of the kinetic energy of the bombarding particle, is equal to 6.057 MeV.

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