$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.

(N/A) The de Broglie wavelength of a particle with mass $m$ and velocity $v$ is given by $\lambda = \frac{h}{mv}$.
For an electron,the wavelength is $\lambda_e = \frac{h}{m_e v_e}$.
Given that the particle's velocity $v = 3v_e$ and the ratio of wavelengths $\frac{\lambda}{\lambda_e} = 1.813 \times 10^{-4}$.
Using the ratio: $\frac{\lambda}{\lambda_e} = \frac{h/mv}{h/m_e v_e} = \frac{m_e v_e}{mv} = 1.813 \times 10^{-4}$.
Rearranging for the mass of the particle $m$: $m = m_e \left( \frac{v_e}{v} \right) \left( \frac{\lambda_e}{\lambda} \right)$.
Substituting the values $m_e = 9.11 \times 10^{-31} \ kg$,$\frac{v_e}{v} = \frac{1}{3}$,and $\frac{\lambda_e}{\lambda} = \frac{1}{1.813 \times 10^{-4}}$:
$m = (9.11 \times 10^{-31} \ kg) \times \left( \frac{1}{3} \right) \times \left( \frac{1}{1.813 \times 10^{-4}} \right) \approx 1.675 \times 10^{-27} \ kg$.
This mass corresponds to the mass of a proton or a neutron.