According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\pi$ (or $h/2\pi$, where $h$ is the Planck's constant). This can be expressed as:

$$ L = n \frac{h}{2\pi} $$

where $n$ is the principal quantum number or the orbit number.

So, for the first orbit ($n=1$), the angular momentum $L_1$ is:

$$ L_1 = 1 \times \frac{h}{2\pi} = \frac{h}{2\pi} $$

And for the second orbit ($n=2$), the angular momentum $L_2$ is:

$$ L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} $$

The change in angular momentum when moving from the first to the second orbit is the difference between $L_2$ and $L_1$:

$$ \Delta L = L_2 - L_1 = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{2\pi} $$

Since $\frac{h}{2\pi}$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.

Therefore, the correct answer is $L$.