Molecule AB has a bond length of 1.617 $$\mathring{A}$$ and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is: $$(e_0 = 4.802 \times 10^{-10}$$ esu$$)$$

We know that the electric dipole moment $$\mu$$ of a diatomic molecule is given by the simple relation

$$\mu = q \, r$$

where

$$q$$ = magnitude of the charge on each atom (in esu),
$$r$$ = bond length (in cm).

The data provided are

Bond length: $$r = 1.617 \,\text{\AA}$$
Dipole moment: $$\mu = 0.38 \,\text{D}$$

First, we must convert the bond length from angstroms to centimeters. Because

$$1 \,\text{\AA} = 10^{-8} \,\text{cm},$$

we have

$$r = 1.617 \times 10^{-8} \,\text{cm}.$$

Next, we convert the dipole moment from debye to the cgs electrostatic unit. By definition,

$$1 \,\text{D} = 10^{-18} \,\text{esu}\,\text{cm},$$

so

$$\mu = 0.38 \times 10^{-18} \,\text{esu}\,\text{cm} = 3.8 \times 10^{-19} \,\text{esu}\,\text{cm}.$$

Now we substitute these values into the formula $$\mu = q r$$ and solve for $$q$$:

$$q \;=\; \frac{\mu}{r} \;=\; \frac{3.8 \times 10^{-19}\,\text{esu}\,\text{cm}}{1.617 \times 10^{-8}\,\text{cm}}.$$

Carrying out the division step by step:

First divide the numerical parts:

$$\frac{3.8}{1.617} \approx 2.35.$$

Next, handle the powers of ten:

$$10^{-19} \div 10^{-8} = 10^{-19 -(-8)} = 10^{-11}.$$

Combining these results, we obtain

$$q \approx 2.35 \times 10^{-11} \,\text{esu}.$$

The question asks for the fractional (or partial) charge on each atom relative to the fundamental electronic charge $$e_0$$. The electronic charge is given as

$$e_0 = 4.802 \times 10^{-10} \,\text{esu}.$$

The fractional charge $$f$$ is therefore

$$f \;=\; \frac{q}{e_0} \;=\; \frac{2.35 \times 10^{-11}\,\text{esu}}{4.802 \times 10^{-10}\,\text{esu}}.$$

Perform the division:

First, divide the coefficients:

$$\frac{2.35}{4.802} \approx 0.489 \times 10^{-1} = 0.0489.$$

(Here, shifting one power of ten from denominator to numerator moves the decimal place one step to the left.)

Hence

$$f \approx 0.05.$$

Thus each atom carries about $$5\%$$ of the elementary charge in magnitude, meaning the fractional charge is $$0.05$$.

Hence, the correct answer is Option B.