Let's approach this problem by understanding the basics and applying the Bohr model to find the energy levels of a hydrogen atom.

The energy of an electron in a hydrogen atom for any given orbit can be defined using the formula:

$$E_n = \frac{E}{n^2}$$

where:

• $E$ is the energy of the electron in the ground state ($n=1$), and
• $n$ is the principal quantum number (orbit number).

The radius of an orbit in the hydrogen atom, according to the Bohr model, is given by:

$$r_n = n^2 a_0$$

where:

• $r_n$ is the radius of the nth orbit,
• $a_0$ is the Bohr radius ($0.529$ angstroms or $\mathop A\limits^o$), and
• $n$ is the principal quantum number (orbit number).

Given that:

• The radius of a certain orbit of the hydrogen atom is $8.48 \mathop A\limits^o$, and
• The energy of an electron in this orbit is $E/x$,
• We need to find $x$.

First, let's find $n$, the principal quantum number for the orbit with radius $8.48$ angstroms:

$$8.48 = n^2 \times 0.529$$

Solving for $n^2$:

$$n^2 = \frac{8.48}{0.529}$$

$$n^2 \approx 16.03$$

For simplicity and practicality in the quantum model, $n^2$ approximately equal to 16 would imply $n = 4$, considering $n$ must be a whole number and $16.03$ is close to $16$, which is a perfect square of $4$.

Now, to find $x$, we'll use the energy relationship. Since the energy levels of the hydrogen atom are inversely proportional to the square of the principal quantum number $n$,

$$E_{orbit} = \frac{E}{n^2} = \frac{E}{4^2} = \frac{E}{16}$$

According to the given information, $E_{orbit} = \frac{E}{x}$, which means:

$$\frac{E}{x} = \frac{E}{16}$$

Hence,

$$x = 16$$