Question 502: Atoms & Nuclei

Question

In the Bohr model, an electron moves in a circular orbit around the nucleus. Considering an orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nth excited state, is
(e = electronic charge, m$$_e$$ = mass of the electron, h = Planck's constant)

Accepted Answer

We need to determine the magnetic moment of the hydrogen atom when the electron is in the nth excited state using the Bohr model. In the Bohr model, an electron in the nth orbit has specific characteristics such as angular momentum and radius. These properties contribute to the overall magnetic moment of the atom.
The magnetic moment $$\mu$$ of a current loop is given by the product of the current $$I$$ and the area $$A$$ enclosed by the loop:
$$ \mu = I \cdot A $$
For a circular loop with radius $$r$$, the area $$A$$ is:
$$ A = \pi r^2 $$
In the case of an electron orbiting in a circle, the current $$I$$ is the charge of the electron $$e$$ divided by the orbital period $$T$$:
$$ I = \frac{e}{T} $$
The orbital period $$T$$ is the time it takes for the electron to complete one orbit, which is the circumference of the orbit divided by the orbital speed $$v$$:
$$ T = \frac{2 \pi r}{v} $$
Thus, the current $$I$$ becomes:
$$ I = \frac{e}{T} = \frac{e}{\frac{2 \pi r}{v}} = \frac{e v}{2 \pi r} $$
Substituting this into the equation for the magnetic moment $$\mu$$, we get:
$$ \mu = \left( \frac{e v}{2 \pi r} \right) \cdot \pi r^2 = \frac{e v r}{2} $$
Now, we use BohrÃ¢ÂÂs quantization condition for angular momentum, which states that the angular momentum $$L$$ is quantized and given by:
$$ m_e v r = n \frac{h}{2 \pi} $$
Solving for $$v r$$, we have:
$$ v r = \frac{n h}{2 \pi m_e} $$
Substituting this back into the expression for the magnetic moment, we obtain:
$$ \mu = \frac{e}{2 m_e} \cdot \left( \frac{n h}{2 \pi} \right) = \left( \frac{e}{2 m_e} \right) \frac{n h}{2 \pi} $$
Therefore, the magnetic moment of the hydrogen atom in the nth excited state is:
$$ \left( \frac{e}{2 m_e} \right) \frac{n h}{2 \pi} $$
This corresponds to option D:
$$ \left( \frac{e}{2 m_e} \right) \frac{n h}{2 \pi} $$

Answer

$$
\left(\frac{\mathrm{e}}{\mathrm{m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}
$$

Answer

$$
\mathrm{\left(\frac{e}{m_e}\right) \frac{n^2 h}{2 \pi}}
$$

Answer

$$
\mathrm{\left(\frac{e}{2 m_e}\right) \frac{n^2 h}{2 \pi}}
$$

Answer

$$
\left(\frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}
$$

Atoms & Nuclei

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