Don't use chat gpt It A Lithium atom collides with a Hydrogen atom and produces two Alphas (He atoms).The reaction is expressed as: 7Li (Z = 3) + ¹H (Z = 1)→ 4He (Z = 2) + 4He (Z = 2) wherethe masses are 7Li (Z = 3): 7.016005 u, ¹H (Z = 1): 1.007825 u, 4He (Z = 2): 4.002603u. The mass difference in the reaction is attributed to the reaction energy which ismost nearly what (Express in MeV)(1 u = 931.5 MeV/c²)O 15.37 MeVO 13.57 MeVO 17.35 MeVO 13.35 MeV

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Answered: A Lithium atom collides with a Hydrogen…

Modern Physics

3rd Edition

ISBN:9781111794378

Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer

Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer

Chapter13: Nuclear Structure

Section: Chapter Questions

Problem 6P

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Question 12 of 22 Calculate the mass defect of Nitrogen (A = 14, Z = 7). The atomic mass of Nitrogen is 14.00307 u. (Note: The mass of a hydrogen atom is mH = 1.007825 u, and the mass of the neutron is mN = 1.008665 u.) 00 1.21 u 0.53 u 0.25 u 0.11 u zero
Use the below values for this problem. Please note that the mass for H is for the entire atom (proton & electron). Neutron: m = 1.67493x10-27 kg = 1.008665 u = 939.57 MeV/c² . ¹H: mH = 1.67353x10-27 kg = 1.007825 u = 938.78 MeV/c² 1 1 u = 1.6605x10-27 kg = 931.5 MeV/c² . Consider the following decay: 239 Pu 235 U+ a. 239 Pu has a mass of 239.0521634 u, 235 U has a mass of 235.0439299 u, and a has a mass of 4.002603 u. 94 92 94 92 Determine the disintegration energy (Q-value) in MeV. Q = Determine the binding energy (in MeV) for 239 Pu. 94 EB =
How many kJ of energy are released to form one mole of 133Cs from protons and neutrons if the atom has a mass of 132.905429 amu? Please remember to include the mass of electrons in the calculation. Given the mass of a proton is 1.007825 amu
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How much energy in kJ is released to form one mole of 75As from protons, electrons, and neutrons if the nucleus has a mass of 74.921594 amu? The masses of the proton, electron and neutron are 1.00728 amu, 0.000549 amu and 1.00867 amu, respectively. -3.15e10 kJ/mol -6.31e10 kJ/mol -6.31e13 kJ/mol 3.15e13 kJ/mol
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Q4. a) Alpha decay is a two body final state process, in which a nucleus 4 AXz →(A−4) Y(z−2) +¹ He₂ + Qa If the masses of the atoms of the parent and the daughter nuclei are (roughly) Au, and (A — 4)u and that of the a particle is 4u, determine the relation between the kinetic energy To of the a particle and Qa. b) Explain why in case of even A nuclei, it is possible to have two most stable isobars, where as for odd A nuclei, there can be only one most stable isobar. c) Construct the mass parabolas for A = 101, and identify the most stable isobars. d) Construct the mass parabolas for A = 106, and identify the two most stable isobars.
The atomic mass of hypothetical element (voll aic) (₁4A) is 4.02 amu. The energy required to dissociate this element into its component parts is: (m₁ =1.009 amu, mp = 1.007amu, neglect the mass of electron)
The element 218Po (Polonium-218, Z = 84, atomic mass 218.008966 u) can decay through the emission of a β-particle to 218At (Astatine-218, Z = 85, atomic mass 218.00880 u). If all of the energy released is carried away by the β-particle's kinetic energy, calculate the kinetic energy of the β-particle.answer in units of MeV, correct to two decimal places.
An Erbium-166 nucleus contains 68 protons. The atomic mass of a neutral Erbium-166 atom is 165.930u, where u = 931.5 MeV/c². In this question you may use that the mass of a proton is 938.27 MeV/c², the mass of a neutron is 939.57 MeV/e² and the mass of an electron is 0.511 MeV/c². i. Calculate the nuclear binding energy per nucleon, giving your answer in units of MeV. ii. Electrons with an energy of 0.5 GeV are scattered off the nucleus. Estimate the scattering angle of the first minimum in the resulting diffraction pattern. iii. Briefly comment on whether or not you expect this nucleus to be spherical, and what consequence this has for excited states of the nucleus in the collective model.
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c) The equation below describes the disintegration of a polonium nucleus into a lead nucleus and an alpha-particle. During the reaction energy Q is released. 210Po → He +²02Pb+Q 84 82 Calculate the loss of energy during the reaction. The masses in the atomic mass unit u are as follows: 210 206 Po= 209.98287 u, Pb = 205.97446 u and He = 4.002604 u. 84 82 You may assume that 1u is equivalent to 931 MeV. d) The lead nucleus recoils in the opposite direction to the emitted alpha particle conserving momentum. Hence calculate: i) The ratio of the recoil nucleus and alpha particle velocities ii) The kinetic energy distribution of these products.

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