3 edited body edited Dec 16 '13 at 21:51 Brandon Enright 10.3k124374 So, I assumed a pair of electron and proton proton to behave as a dipole ClassicalyClassically, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localisedlocalized electron  ,but but about orbitals. This is because particles have an associated wavefunction $$\psi(t)$$, which tells you the probability of finding your particle at a particular place. In the case of an electron and a proton (hydrogen atom), the orbital has spherical symmetry. This means that the will be no net charge. So, I assumed a pair of electron and proton to behave as a dipole Classicaly, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localised electron  ,but about orbitals. This is because particles have an associated wavefunction $$\psi(t)$$, which tells you the probability of finding your particle at a particular place. In the case of an electron and a proton (hydrogen atom), the orbital has spherical symmetry. This means that the will be no net charge. So, I assumed a pair of electron and proton to behave as a dipole Classically, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localized electron, but about orbitals. This is because particles have an associated wavefunction $$\psi(t)$$, which tells you the probability of finding your particle at a particular place. In the case of an electron and a proton (hydrogen atom), the orbital has spherical symmetry. This means that the will be no net charge. 2 added 184 characters in body edited Dec 16 '13 at 21:42 jinawee 9,01543878 So, I assumed a pair of electron and proton to behave as a dipole Classicaly, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localised electron ,but a wavefunctionabout orbitals. This is because particles have an associated wavefunction $$\psi(t)$$, which tells you the probability of finding your particle at a particular place. In the case of an electron and a proton (hydrogen atom), the orbital has spherical symmetry. This means that the will be no net charge. So, I assumed a pair of electron and proton to behave as a dipole Classicaly, this is correct. But in quantum mechanics we don't talk about a localised electron ,but a wavefunction. In the case of an electron and proton (hydrogen), the orbital has spherical symmetry. This means that the will be no net charge. So, I assumed a pair of electron and proton to behave as a dipole Classicaly, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localised electron ,but about orbitals. This is because particles have an associated wavefunction $$\psi(t)$$, which tells you the probability of finding your particle at a particular place. In the case of an electron and a proton (hydrogen atom), the orbital has spherical symmetry. This means that the will be no net charge. 1 answered Dec 16 '13 at 21:37 jinawee 9,01543878 So, I assumed a pair of electron and proton to behave as a dipole Classicaly, this is correct. But in quantum mechanics we don't talk about a localised electron ,but a wavefunction. In the case of an electron and proton (hydrogen), the orbital has spherical symmetry. This means that the will be no net charge.