Virtual photons are particles that mediate the Coulomb force and travel at the speed of light. What is the approximate energy of the virtual photons that mediate between the ground state electron and the proton in a hydrogen atom? The radius of the electron in the ground state is 0.053 nm.
A) 100 eV
B) 500 eV
C) 1000 eV
D) 4 keV
E) 0.5 MeV

Question

Quizplus · Accepted Answer

The approximate energy of the Coulomb force between the electron and proton in a hydrogen atom is given by:
$E = \frac{e^2}{4\pi\epsilon_0r} = \frac{(1.60 	imes 10^{-19} C)^2}{4\pi(8.85	imes 10^{-12} C^2/Nm^2)(0.053 	imes 10^{-9} m)} = 13.6 eV$
This energy corresponds to the energy of a photon that would be emitted if the electron were to jump from the ground state to the ionized state. However, the virtual photons that mediate the Coulomb force do not have this exact energy, as they are not real particles that can be detected. They only exist as a mathematical construct to explain the interaction between charged particles. Therefore, there is no exact value for the energy of the virtual photons. However, we can estimate their energy by assuming that they have a wavelength on the order of the radius of the electron in the ground state. Using the de Broglie wavelength, we can calculate the momentum and energy of a photon with this wavelength:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_eE}}$
where $m_e$ is the mass of the electron and $E$ is the energy of the photon. Solving for $E$, we get:
$E = \frac{h^2}{2m_e\lambda^2} = \frac{(6.63 	imes 10^{-34} J\cdot s)^2}{2(9.11 	imes 10^{-31} kg)(0.053 	imes 10^{-9} m)^2} = 4.17 keV$
This represents an upper limit on the energy of the virtual photons, as the actual wavelength of the photons is likely shorter than the electron radius. Therefore, the best answer is D) 4 keV, which is the closest choice to our estimate of the virtual photon energy.

Virtual Photons Are Particles That Mediate the Coulomb Force and Travel