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Question 21 5 pts Calculate the kinetic energy (in MeV) of an alpha particle whose DeBroglie...
please answer all questions What is the DeBroglie wavelength of an antineutrino with a kinetic energy...
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 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...
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 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 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...
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...
can you answer all the questions posted A soap film (n = 1.33) sits on top...
can
you answer all the questions posted
A soap film (n = 1.33) sits on top of a plastic plate (n = 1.37). White light strikes the film nearly perpendicularly. The smallest nonzero film thickness for which the film appears bright due to constructive interference is 153 nm. What color does the film appear? O Violet (420 nm) O Yellow (580 nm Red (650 nm) Blue (475 nm O Green (500 m) A Ba-135 atom (barium-135) has a nucleus with...
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...
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 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...
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 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 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...
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...
can
you answer all the questions posted
A soap film (n = 1.33) sits on top of a plastic plate (n = 1.37). White light strikes the film nearly perpendicularly. The smallest nonzero film thickness for which the film appears bright due to constructive interference is 153 nm. What color does the film appear? O Violet (420 nm) O Yellow (580 nm Red (650 nm) Blue (475 nm O Green (500 m) A Ba-135 atom (barium-135) has a nucleus with...
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...
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