**Explanation:**

To determine the orbit in which the emission Be3 is placed, we need to analyze the wavelength of the observed emission and relate it to the energy levels of the electron in a hydrogen-like atom.

**1. Wavelength of the observed emission:**
The given wavelength of the observed emission is 2.116 Å.

**2. Relationship between wavelength and energy levels:**
According to the Bohr model of the hydrogen-like atom, the wavelength of the emitted photon is related to the energy levels of the electron by the Rydberg formula:

1/λ = R * (1/n1^2 - 1/n2^2)

where:
- λ is the wavelength of the emitted photon,
- R is the Rydberg constant,
- n1 and n2 are the principal quantum numbers of the initial and final energy levels, respectively.

**3. Calculating the energy levels:**
To determine the energy levels, we need to rearrange the Rydberg formula:

1/n2^2 = 1/n1^2 - 1/λR

Since Be3 is a hydrogen-like atom, we can assume that the initial energy level is the ground state (n1 = 1). Plugging in the values, we get:

1/n2^2 = 1/1^2 - 1/(2.116 A * R)

**4. Determining the orbit:**
To find the orbit (principal quantum number), we need to solve for n2. Rearranging the equation, we have:

1/n2^2 = 1 - (2.116 A * R)

Simplifying, we find:

1/n2^2 = 1 - (2.116 * 10^-10 m * R)

Comparing this equation with the general form of the Rydberg formula:

1/n^2 = 1 - (1/n1^2 - 1/n2^2)

We can conclude that the principal quantum number n is equal to n2. Therefore, the emission Be3 is placed in the fourth orbit (n = 4).