Magnetic Field Ratio Calculation

This problem requires calculating the ratio of magnetic fields generated by electron motion in different Bohr orbits of a hydrogen atom.

Bohr Model Principles

• In Bohr's model, the electron orbits are quantized.
• The magnetic field ($B$) generated by the orbiting electron is proportional to the current ($I$) and inversely proportional to the orbit's radius ($r$): $B \propto \frac{I}{r}$.
• The current is $I = \frac{e}{T}$, where $e$ is the elementary charge and $T$ is the time period.
• The time period $T = \frac{2\pi r}{v}$, where $v$ is the electron's speed.
• Thus, $I = \frac{ev}{2\pi r}$.
• Substituting $I$ into the magnetic field expression: $B \propto \frac{ev/2\pi r}{r} = \frac{ev}{2\pi r^2}$.

Orbit Properties in Bohr Model

For a hydrogen atom in Bohr's model:

• Electron speed $v \propto \frac{1}{n}$, where $n$ is the principal quantum number.
• Orbit radius $r \propto n^2$.

Magnetic Field Dependence on Quantum Number

Substitute the dependencies of $v$ and $r$ into the expression for $B$:

$B \propto \frac{e \cdot (1/n)}{2\pi \cdot (n^2)^2} = \frac{e}{2\pi} \cdot \frac{1}{n \cdot n^4} = \frac{e}{2\pi} \cdot \frac{1}{n^5}$

This shows that the magnetic field $B$ is inversely proportional to the fifth power of the principal quantum number: $B \propto \frac{1}{n^5}$.

Calculating the Ratio

We need to find the ratio $\frac{B_2}{B_4}$ for the 2nd ($n=2$) and 4th ($n=4$) orbits.

Using the proportionality $B \propto \frac{1}{n^5}$:

$ \frac{B_2}{B_4} = \frac{1/n_2^5}{1/n_4^5} = \frac{n_4^5}{n_2^5} $

Substitute $n_2 = 2$ and $n_4 = 4$:

$ \frac{B_2}{B_4} = \frac{4^5}{2^5} = \left(\frac{4}{2}\right)^5 = 2^5 $

$ 2^5 = 32 $

Conclusion

The ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits is 32.