54P. - University Physics with Modern Physics

Question

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed of 6 2.2 10 m/s in a circular orbit of radius 5.3 10 m. (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current I? (c) What is the magnetic moment of the atom due to the motion of the electron?

Step-by-Step Solution

Answer

1. The orbital period of the electron is \(T = 1.5 \times {10^{ - 16}}\,{\rm{s}}\)
2. The current is \(I = 1.1 \times {10^{ - 3}}\,{\rm{A}}\)
3. The magnetic moment is \(\mu  = 9.71 \times {10^{ - 24}}\,{\rm{A}}{{\rm{m}}^2}\)

Step 1: The significance of the time period, current and magnetic moment

Time period of electron is given by

\(T = \frac{{2\pi r}}{v}\)

The current is given by

\(I = \frac{Q}{t}\)

The magnetic moment is given by

\(\mu  = IA = I\pi {r^2}\)

Step 2: Identification of the given data

The speed of proton is, v = 2.2 10 m/s 

The radius of the orbit is, r = 5.3 10 m 

Step 3: Determine the time period

(a)

 

Orbital period of an electron is given by

\(T = \frac{{2\pi r}}{v}\)

 

Substitute all the value in the above equation.

\(\begin{aligned}T = \frac{{2\pi \left( {5.3 \times {{10}^{ - 11}}\,{\rm{m}}} \right)}}{{2.2 \times {{10}^6}\,{\rm{m/s}}}}\\ = 1.5 \times {10^{ - 16}}\,{\rm{s}}\end{aligned}\)

 

Therefore, the orbital period of the electron is \(T = 1.5 \times {10^{ - 16}}\,{\rm{s}}\)

Step 4: Determine the current

(b)

 

The current is given by

\(\begin{aligned}I = \frac{Q}{t}\\I = \frac{e}{t}\end{aligned}\)

 

Substitute all the value in the above equation.

\(\begin{aligned}I = \frac{{1.6 \times {{10}^{ - 19}}\,{\rm{C}}}}{{1.5 \times {{10}^{ - 16}}\,{\rm{s}}}}\\ = 1.1 \times {10^{ - 3}}\,{\rm{A}}\end{aligned}\)

 

Therefore, the current is \(I = 1.1 \times {10^{ - 3}}\,{\rm{A}}\)

Step 5: Determine the magnetic moment

(c)

 

The magnetic moment is given by

\(\begin{aligned}\mu  = IA\\\mu   = I\pi {r^2}\end{aligned}\)

 

Substitute all the value in the above equation.

\(\begin{aligned}\mu  = \left( {1.1 \times {{10}^{ - 3}}\;{\rm{A}}} \right)\pi {\left( {5.3 \times {{10}^{ - 11}}\;{\rm{m}}} \right)^2}\\ = 9.71 \times {10^{ - 24}}\;{\rm{A}}{{\rm{m}}^2}\end{aligned}\)

 

Therefore, the magnetic moment is \(\mu  = 9.71 \times {10^{ - 24}}\;{\rm{A}}{{\rm{m}}^2}\)