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 × 10–11 m . Calculate the magnetic field strength at the proton due to the orbital motion of the electron in the n = 1 state.

Respuesta :

Answer:

B = 12.46 T

Explanation:

At n = 1 state we know that radius is given as

[tex]R = 5.29 \times 10^{-11} m[/tex]

now we have

[tex]T = \frac{2\pi R}{v}[/tex]

here we know that speed is given in that

[tex]v = 2.18 \times 10^6 m/s[/tex]

now the time period is given as

[tex]T = \frac{2\pi R}{v}[/tex]

[tex]T = \frac{2\pi (5.29 \times 10^{-11})}{2.18 \times 10^6}[/tex]

[tex]T = 1.52 \times 10^{-16} s[/tex]

Now the electric current due to revolution of charge is given by

[tex]i = \frac{e}{T}[/tex]

[tex]i = \frac{1.6 \times 10^{-19}}{1.52 \times 10^{-16}}[/tex]

[tex]i = 1.05 \times 10^{-3} A[/tex]

now magnetic field at the center position is given as

[tex]B = \frac{\mu_0 i}{2R}[/tex]

[tex]B = \frac{4\pi \times 10^{-7} (1.05 \times 10^{-3})}{2(5.29 \times 10^{-11}}[/tex]

[tex]B = 12.46 T[/tex]