Actually this is a very basic question about high frequency (HF) applications.
If one is talking about HF, the wavelength is short in comparison to the transmission line, which is the case in an antenna.
The wavelength is given by
$$ \lambda = \frac{c}{f}\,.$$
The frequency \$f\$ is the operating frequency.
The propagation speed \$c\$ is given by
$$ c= \frac{1}{\epsilon_0\epsilon\cdot\mu_0\mu}\,.$$
If we have free space, \$\epsilon\$ and \$ \mu\$ are one and \$c\$ simplifies to the speed of light \$c_0\$. If there is dielectric material, in whicht a wave propagates, the speed of the wave is dependent on these two parameters (\$\epsilon\$, \$ \mu\$).
So the quantity of how the wave is propagating is important in the design of an antenna.
With the wavelength \$\lambda\$, one can directly see, which dimensions the antenna has, without further calculations.
\$I\$ shows the current distribution and \$V\$ the voltage distribution over a half-wave dipole antenna. So one can see that \$\lambda=2\pi\$, where \$2\pi\$ is one period of the voltage/current distribution. The current distribution over the dipole plays an important role e.q. for the radiation pattern. So it makes sense to give the quantity \$\lambda\$ to directly have a feeling for the current distribution over the piece of wire.
Furthermore, the wavelength is just a length, so its quantity is given in SI units, for example meter
$$[\lambda] = \mathrm{m} \,.$$
