In a simple picture of the hydrogen atom,the electron moves in circular orbits around the central proton attracted by the Coulomb force.The lowest (n = 1)energy orbit that is allowed for the electron is at a radius of 5.29 $\times$ 10^-11 m and the second (n = 2)lowest allowed orbit is at 4 times this radius.Calculate the magnetic field strength at the proton due to the orbital motion of the electron in the n = 2 state. A)0 T B)6.3 T C)0.39 T D)12.5 T E)None of the above

Question

Quizplus · Accepted Answer

The magnetic field strength at the proton due to the orbital motion of the electron is given by:

B = μ0ev/2r^2

Where μ0 is the permeability of free space, e is the charge of the electron, v is the velocity of the electron in its orbit, and r is the radius of the orbit.

For the n = 2 state, the radius is 4 times larger than the n = 1 state, so the magnetic field strength is decreased by a factor of 16.

Therefore, the magnetic field strength at the proton due to the orbital motion of the electron in the n = 2 state is:

B = μ0ev/2(4r)^2 = μ0ev/32r^2

Substituting the values given (e = 1.6 × 10^-19 C, r = 4 × 5.29 × 10^-11 m, v = e/mvr) and simplifying, we get:

B = 0.39 T

Therefore, the answer is C.

In a Simple Picture of the Hydrogen Atom,the Electron Moves ×\times×

In a Simple Picture of the Hydrogen Atom,the Electron Moves $\times$