When a proton moves freely in a magnetic field, its momentum will not change but its direction will be affected. The proton will experience a force perpendicular to its velocity, causing it to move in a circular path, hence its speed will remain constant.
The spinning proton magnetic field influences the behavior of particles in a magnetic field by aligning with or against the external magnetic field. This alignment affects the energy levels of the particles, leading to phenomena like nuclear magnetic resonance and the Zeeman effect.
The velocity of a proton in a magnetic field depends on the strength of the magnetic field and the charge of the proton. The velocity can be calculated using the formula v (qB) / m, where v is the velocity, q is the charge of the proton, B is the strength of the magnetic field, and m is the mass of the proton.
A proton with high energy and momentum is required to initiate a nuclear reaction. When this high-energy proton collides with another nucleus, it can cause it to undergo fission or fusion, releasing a large amount of energy.
An electric field will exert a force on a proton due to its positive charge. The proton will experience a force in the direction of the electric field if the field is uniform, causing it to accelerate in that direction.
mass of the proton is 0. Answer 2 But the question asked about photons, not protons. The mass of a photon is also 0, though the mass of a proton is not!
The spinning proton magnetic field influences the behavior of particles in a magnetic field by aligning with or against the external magnetic field. This alignment affects the energy levels of the particles, leading to phenomena like nuclear magnetic resonance and the Zeeman effect.
The velocity of a proton in a magnetic field depends on the strength of the magnetic field and the charge of the proton. The velocity can be calculated using the formula v (qB) / m, where v is the velocity, q is the charge of the proton, B is the strength of the magnetic field, and m is the mass of the proton.
The electron, because it has much less mass than the proton and momentum is the product of mass and speed.
No. Electron is roughly 1/2000 of proton in mass. If the question is about spacial size, it is not a meaningful question. The idea that elementary particles are like a physical ping pong ball (but smaller) is incorrect. Elementary particles are not similar to the every day object except in the fact, that they possess energy and momentum, can have electrical charge, magnetic moment and angular momentum.
No, it is not possible ,a proton is the combination of two up and one down quarks but quarks do not exist freely.
the electron would have the longer wavelength b/c the proton has more momentum and λ=h/p (λ is wavelength, h is planc's constant and p is momentum)
Use the left hand rule (or use the right hand rule and reverse the result):F(vector) = v(vector) x B(vector)v is up and B is into page, so F is to the right. ?But since the electron still has momentum moving up, it will start moving up and to the right (assuming the proton is fixed. ?If the proton is also moving, then it gets more complicated).
Nils Myrberg has written: 'Proton magnetic resonance in human dental enamel and dentine' -- subject(s): Dental enamel, Dentin, Nuclear magnetic resonance, Proton magnetic resonance, Spectra, Spectrum analysis
Barry Blumenfeld has written: 'Observation of pi- mesons with large transverse momentum in high energy proton-proton collisions' -- subject(s): Collisions (Nuclear physics), Proton-proton interactions, Protons, Scattering
A proton is surrounded by an electric field, which interacts with other charged particles. It also interacts with a magnetic field under certain conditions, such as when it moves through a magnetic field.
Hydrogen has a magnetic moment due to its single proton in the nucleus. This makes hydrogen weakly magnetic, but its magnetic properties are not significant for most practical purposes.
When alkynyl molecules are placed in NMR instrument the induced magnetic field of molecules are in Diamagnetic region of external magnetic field. There fore the resultant energy will be low