Answer
Proton
Work Step by Step
If $m_e$ and $v_e$ the mass and velocity of the electron. The de Broglie wavelength of the electron is given by
$\lambda_e=\frac{h}{m_ev_e}$
Let $m$ and $v$ the mass and velocity of the unknown particle. The de Broglie wavelength of the unknown particle is given by
$\lambda=\frac{h}{mv}$
Given $v=3v_e$, Then, $\lambda=\frac{h}{3mv_e}$
The ratio of the de Broglie wavelength of the particle to that of the electron is given by
$\frac{\lambda}{\lambda_e}=1.813\times 10^{-4}$
or, $\frac{\frac{h}{3mv_e}}{\frac{h}{m_ev_e}}=1.813\times 10^{-4}$
or, $\frac{m_e}{3m}=1.813\times 10^{-4}$
or, $m=\frac{m_e}{3\times 1.813\times 10^{-4}}\approx 1839m_e$
Therefore, the mass of the particle is ~$1839$ times the mass of an electron, which indicates that the particle is proton.
