Question

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.

Answer 1

B = 12.46 T

Step-by-step explanation:

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

`R = 5.29 * 10^(-11) m`

now we have

`T = (2\pi R)/(v)`

here we know that speed is given in that

`v = 2.18 * 10^6 m/s`

now the time period is given as

`T = (2\pi R)/(v)`

`T = (2\pi (5.29 * 10^(-11)))/(2.18 * 10^6)`

`T = 1.52 * 10^(-16) s`

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

`i = (e)/(T)`

`i = (1.6 * 10^(-19))/(1.52 * 10^(-16))`

`i = 1.05 * 10^(-3) A`

now magnetic field at the center position is given as

`B = (\mu_0 i)/(2R)`

`B = (4\pi * 10^(-7) (1.05 * 10^(-3)))/(2(5.29 * 10^(-11))`

`B = 12.46 T`