The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :

(Assume $m_{p}=m_{e} \times 1849$ )

The de Broglie wavelength (λ) of a particle can be found using the formula: <br/><br/> $\lambda = \frac{h}{p}$ <br/><br/> 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: <br/><br/> $p = \sqrt{2mK}$ <br/><br/> Combining these two equations, we get: <br/><br/> $\lambda = \frac{h}{\sqrt{2mK}}$ <br/><br/> 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: <br/><br/> $m_p = 1849 \times m_e$ <br/><br/> Let's find the ratio of the de Broglie wavelengths of the proton ($λ_p$) and the electron ($λ_e$): <br/><br/> $$ \frac{\lambda_p}{\lambda_e} = \frac{\frac{h}{\sqrt{2m_pK}}}{\frac{h}{\sqrt{2m_eK}}} $$ <br/><br/> Simplifying the expression, we get: <br/><br/> $\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2m_eK}}{\sqrt{2m_pK}}$ The 2K terms cancel out: <br/><br/> $\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}}$ <br/><br/> Substitute the given relationship between the masses: <br/><br/> $\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{1849 \times m_e}}$ <br/><br/> Further simplification: <br/><br/> $\frac{\lambda_p}{\lambda_e} = \frac{1}{\sqrt{1849}}$ <br/><br/> Since 1849 is equal to $43^2$: <br/><br/> $\frac{\lambda_p}{\lambda_e} = \frac{1}{43}$ <br/><br/> Thus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.