The de Broglie wavelength (λ) of a particle can be found using the formula:

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

where h is the Planck constant and p is the momentum of the particle. The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:

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

Combining these two equations, we get:

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

Now, we are given that the kinetic energy of the proton and electron is the same. Let's denote the masses of the proton and electron as $$m_p$$ and $$m_e$$, respectively. We are given the relationship between the two masses:

$$ m_p = 1849 \times m_e $$

Let's find the ratio of the de Broglie wavelengths of the proton ($λ_p$) and the electron ($λ_e$):

$$ \frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_pK}}}{\frac{h}{\sqrt{2m_eK}}} $$

Simplifying the expression, we get:

$$ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2m_eK}}{\sqrt{2m_pK}} $$ The 2K terms cancel out:

$$ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}} $$

Substitute the given relationship between the masses:

$$ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{1849 \times m_e}} $$

Further simplification:

$$ \frac{\lambda_p}{\lambda_e} = \frac{1}{\sqrt{1849}} $$

Since 1849 is equal to $$43^2$$:

$$ \frac{\lambda_p}{\lambda_e} = \frac{1}{43} $$

Thus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.