t 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 Bohr’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:

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

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