Question

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circle of radius 5.7 × 10-11 m. Considering the orbiting electron to be a small current loop, determine the magnetic moment associated with this motion. (Hint: The electron travels around the circle in a time equal to the period of the motion.)

Answer 1

`1.5048* 10^(-23)\ Am^2`

Step-by-step explanation:

q = Charge of proton =
`1.6* 10^(-19)\ C`

r = Radius of circle =
`5.7* 10^(-11)\ m`

v = Velocity of proton =
`3.3* 10^6\ m/s`

Magnetic moment is given by

`M=(1)/(2)qrv\\\Rightarrow M=(1)/(2)1.6* 10^(-19)* 5.7* 10^(-11)* 3.3* 10^6\\\Rightarrow M=1.5048* 10^(-23)\ Am^2`

The magnetic moment associated with this motion is
`1.5048* 10^(-23)\ Am^2`