http://www.antenna-theory.com/antennas/shortdipole.php is a website with useful info., including formulas.
To oversimplify, it seems to say that once the antenna is a tenth or less of the wavelength, the exact ratios don't matter so much. The antenna is inefficient, but it works for both sending and receiving. If you can detect the signal, of course you can amplify it as much as you want.
The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength. It is characterized in the table as a point particle with mass zero and spin one. Its energy is given by $E=h\nu$, where $\nu$ is the frequency of the classical electromagnetic wave which can be built up by photons of the same energy.
This is where the confusion comes. The wavelength and frequency characterize the emergent electromagnetic wave from very many photons. How the classical wave emerges can be seen here although it needs a quantum field theory background to understand it. The photon, as a quantum mechanical entity, has a quantum mechanical wavefunction. This wavefunction complex conjugate squared gives the probability density for the specific photon to be at $(x,y,z,t)$. The frequency in the wavefunction is the frequency of the possible emergent classical wave, but for the individual photon it is only connected with probability of manifestation, as for example in the single photon double slit experiments.
single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames
you ask:
"How can a photon have a wavelength of a few kilometres and yet still be thought of as a particle?
It does not. It takes zillions of photons to build up the classical electromagnetic wave. In the photos above each individual photon gives a little dot. The build up gives the probability density distribution for photons, and lo, there is a frequency associated with the interference pattern, even though the photon manifests individually as a dot at the $(x,y)$ of the screen.
That is why we need quantum mechanics.
Edit after this question became the main duplicate of another one, where I have a long answer/comment that might be of interest to readers.
Does a single photon have a wavelength or not? [duplicate]
Best Answer
Photon frequency and wavelength are the same as the corresponding classical mode.
When you quantize the electromagnetic field, you first treat the spatial dependence by decomposing the field into normal modes, which is a generalization of treating fields of the form $$E(t)=E_0(t)\sin(kx),$$ like you'd find in a conducting box of length $L$ such that $k=\pi n/L$ for some integer $n$. This is really no more than kinematics, or a restatetement of the problem if you will. The true dynamics of the field are then encoded in the temporal dependence. This dependence is through $E_0$, which obeys Maxwell's equations in the form $$\frac{d^2E_0}{dt^2}+\omega^2 E_0=0$$ for $\omega=ck$.
This equation describes a harmonic oscillator, and quantum mechanics says that harmonic-oscillator systems can only have a discrete set of possible energies, with an even spacing of $\hbar\omega$ between them. If the state of the field is such that there is, whenever you look, only one excitation present, then we say the field is in a single-mode, single photon state. This photon then has a well-defined frequency ($\nu=\omega/2\pi$) and wavelength ($\lambda=2\pi/k$).
There can then be another additional complication. In any given box, and more so in free space, there will always be more than one mode present. There is then the possibility that there be exactly one photon in the system, but that we don't know in which mode it's in - the photon is in a Schrödinger's-cat state. Usually the photon will be concentrated in modes within some bandwidth $\Delta\omega$ around some central frequency $\omega_0$, which means that the photon frequency is uncertain to some extent. The remarkable thing is that the electric field will then be nonzero in some spatially and temporally localized region: it will pass any given point in time of order $1/\Delta\omega$ and thus have a width of order $c/\Delta\omega$.
It is also important to realize that you can get this same effect with a classical field: a localized wavepacket will have a corresponding spread in its frequency and wavelength - exactly the same as it's hard to tell the pitch of short notes. The difference between a one-photon state and a weak classical field is a statistical one: for the same mean intensity, a one-photon wavepacket will always give one count on a photo-multiplier tube or avalanche photodiode, whereas a classical field will sometimes give none, sometimes several. This seems like a small difference but it makes things like single-photon interferometry possible.