The de Broglie wavelength of a particle can be expressed as:

$$ \lambda = \frac{h}{p} $$

where:

• $$ h $$ is Planck's constant, and
• $$ p $$ is the momentum of the particle.

For an electron accelerated through a potential difference of $$ V $$ volts, its kinetic energy $$ K $$ is given by:

$$ K = eV $$

where $$ e $$ is the charge of an electron.

The electron's momentum can be expressed in terms of its kinetic energy as:

$$ p = \sqrt{2mK} $$

where $$ m $$ is the mass of the electron.

Substituting this into the de Broglie wavelength equation, we get:

$$ \lambda = \frac{h}{\sqrt{2mK}} $$

Comparing this with the given equation $$ \lambda = \frac{1.227}{x} \, nm $$, we see that $$ x $$ must correspond to:

$$ x = \sqrt{2mK} $$

Since $$ K = eV $$, we can substitute this into our expression for $$ x $$ to get:

$$ x = \sqrt{2meV} = \sqrt{V} $$

where we've used the fact that the mass and charge of an electron are constants.