To find the de Broglie wavelength associated with an alpha particle emitted with a kinetic energy of 1 MeV, we can use the de Broglie wavelength formula:

\[\lambda = \frac{h}{p}\]

where \(\lambda\) is the wavelength, \({h}\) is Planck’s constant (\(6.626 \times 10^{-34} \text{ J s}\)), and \({p}\) is the momentum of the particle.

The momentum \({p}\) can be expressed in terms of kinetic energy \({E}\) for non-relativistic speeds as:

\[p = \sqrt{2mE}\]

where \({m}\) is the mass of the alpha particle (approximately \(6.64 \times 10^{-27} \text{ kg}\)) and \({E}\) is the kinetic energy converted to joules from MeV. Given \(E = 1 \text{ MeV}\), we convert this to joules:

\[1 \text{ MeV} = 1.602 \times 10^{-13} \text{ Joules}\]

Substituting these values into the expression for momentum:

\[p = \sqrt{2 \times 6.64 \times 10^{-27} \text{ kg} \times 1.602 \times 10^{-13} \text{ J}}\]

Calculating the above expression:

\[p = \sqrt{2 \times 6.64 \times 1.602 \times 10^{-40}} \approx 1.45 \times 10^{-22} \text{ kg m/s}\]

Now, substituting the values of \({h}\) and \({p}\) in the de Broglie wavelength formula:

\[\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{1.45 \times 10^{-22} \text{ kg m/s}} \approx 4.57 \times 10^{-12} \text{ m}\]

Converting meters to centimeters (\(1 \text{ m} = 100 \text{ cm}\)):

\[\lambda \approx 4.57 \times 10^{-10} \text{ cm}\]

The closest option to this calculation (after converting from meters to centimeters for comparison) is \(10^{-12} \text{ cm}\).

Thus, the de Broglie wavelength associated with an alpha particle emitted with a kinetic energy of 1 MeV by the nucleus of an atom of radon is approximately of the order of \(10^{-12} \text{ cm}\).