Deuterium resembles hydrogen in properties but:

We recall that deuterium, symbolised as $$^2\text{H}$$ or $$\text{D}$$, is an isotope of ordinary hydrogen, written as $$^1\text{H}$$. Both isotopes possess one proton in the nucleus, so their chemical properties, which largely depend on electronic configuration, are almost identical. However, there is an important difference: deuterium contains one neutron in addition to the single proton, whereas protium (ordinary hydrogen) does not.

This extra neutron doubles the mass of the nucleus. Consequently, the atomic mass of deuterium is $$2\,\text{u}$$ while that of protium is $$1\,\text{u}$$. The heavier mass influences the velocity of the atoms or molecules in a gas at a given temperature. According to the kinetic-theory relation $$\dfrac{1}{2}m v^2 = \dfrac{3}{2}kT,$$ where $$m$$ is the mass of a particle, $$v$$ its root-mean-square speed, $$k$$ Boltzmann’s constant and $$T$$ absolute temperature, we observe that for a fixed $$T$$ the speed $$v$$ is inversely proportional to the square root of the mass $$m$$:

$$v \propto \dfrac{1}{\sqrt{m}}.$$

Substituting $$m = 1\,\text{u}$$ for protium and $$m = 2\,\text{u}$$ for deuterium, we immediately see that

$$\dfrac{v_{\text{D}}}{v_{\text{H}}} = \sqrt{\dfrac{m_{\text{H}}}{m_{\text{D}}}} = \sqrt{\dfrac{1}{2}} = \dfrac{1}{\sqrt{2}} \lt 1.$$

Thus deuterium atoms move more slowly than ordinary hydrogen atoms at the same temperature. Reaction rates for gases depend on collision frequency, which in turn depends on molecular speed. Because $$v_{\text{D}}$$ is smaller, the number of effective collisions per unit time is reduced. Therefore reactions involving deuterium proceed more slowly than analogous reactions with protium. This phenomenon is known as the “isotope effect”.

Now let us match this reasoning with the given options:

A. “reacts vigorously than hydrogen” - contradicts the isotope effect.
B. “reacts just as hydrogen” - ignores the mass-dependent rate difference.
C. “emits $$\beta^+$$ particles” - refers to radioactive decay, but deuterium is stable and does not undergo $$\beta^+$$ emission.
D. “reacts slower than hydrogen” - exactly expresses the isotope effect we have explained.

Hence, the correct answer is Option D.