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please answer all questions What is the DeBroglie wavelength of an antineutrino with a kinetic energy...
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What is the DeBroglie wavelength of an antineutrino with a kinetic energy of 1.8 MeV? Take the mass of the antineutrino to be 4.99E-37 kg. O 1.75 nm 2.88E-4 nm O 5.36E-16 nm 8.45E-12 nm O 1.24 nm The electronic configuration for boron is 1s 2s22p1, and the electronic configuration for aluminum is 152 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy...
Question 21 5 pts Calculate the kinetic energy (in MeV) of an alpha particle whose DeBroglie...
Question 21 5 pts Calculate the kinetic energy (in MeV) of an alpha particle whose DeBroglie wavelength is equal to 1 femtometer (1E-15 m), which is about the size of a nucleus. The mass of the alpha particle is 6.65E-27 kg. 3.3E-11 MeV 5.3E-30 MeV 49.8 MeV 103 MeV 206 MeV Question 22 5 pts The electronic configuration for boron is 1s 2s22p1, and the electronic configuration for aluminum is 1s22s22p6 352 3p1. It requires 8.298 eV to remove the...
The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2...
The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be equal to...
The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s22p6 352...
The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a porbital, the energy required will be equal to 8.298 ev. Since...
0.747 m/s Question 22 5 The electronic configuration for boron is 1s22s22p1, and the electronic configuration...
0.747 m/s Question 22 5 The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will...
Question 22 5 pts The electronic configuration for boron is 152 2s22p1, and the electronic configuration...
Question 22 5 pts The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy...
Question 22 5 pts The electronic configuration for boron is 1s2 2s22p1, and the electronic configuration...
Question 22 5 pts The electronic configuration for boron is 1s2 2s22p1, and the electronic configuration for aluminum is 152 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p 1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the...
The electronic configuration for boron is 152 252 2p1, and the electronic configuration for aluminum is...
The electronic configuration for boron is 152 252 2p1, and the electronic configuration for aluminum is 192 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be...
Question 22 5 pts The electronic configuration for boron is 1s22s22p1, and the electronic configuration for...
Question 22 5 pts The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s2 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (21) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 ev. Since each electron is in ap orbital, the energy required will be...
The electronic configuration for boron is 1s2 2s2 2p1, and the electronic configuration for aluminum is...
The electronic configuration for boron is 1s2 2s2 2p1, and the electronic configuration for aluminum is 1s2 2s2 2p6 3s2 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be equal...
please answer all questions
What is the DeBroglie wavelength of an antineutrino with a kinetic energy of 1.8 MeV? Take the mass of the antineutrino to be 4.99E-37 kg. O 1.75 nm 2.88E-4 nm O 5.36E-16 nm 8.45E-12 nm O 1.24 nm The electronic configuration for boron is 1s 2s22p1, and the electronic configuration for aluminum is 152 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy...
Question 21 5 pts Calculate the kinetic energy (in MeV) of an alpha particle whose DeBroglie wavelength is equal to 1 femtometer (1E-15 m), which is about the size of a nucleus. The mass of the alpha particle is 6.65E-27 kg. 3.3E-11 MeV 5.3E-30 MeV 49.8 MeV 103 MeV 206 MeV Question 22 5 pts The electronic configuration for boron is 1s 2s22p1, and the electronic configuration for aluminum is 1s22s22p6 352 3p1. It requires 8.298 eV to remove the...
The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be equal to...
The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a porbital, the energy required will be equal to 8.298 ev. Since...
0.747 m/s Question 22 5 The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will...
Question 22 5 pts The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy...
Question 22 5 pts The electronic configuration for boron is 1s2 2s22p1, and the electronic configuration for aluminum is 152 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p 1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the...
The electronic configuration for boron is 152 252 2p1, and the electronic configuration for aluminum is 192 252 2p6 352 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV. Since each electron is in a p orbital, the energy required will be...
Question 22 5 pts The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s2 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (21) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)? Since each is the outermost electron around the atom, the energy required will be equal to 8.298 ev. Since each electron is in ap orbital, the energy required will be...
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