A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is:

The de Broglie wavelength of a particle is given by $$\lambda = \frac{h}{mv}$$, where $$h$$ is Planck's constant, $$m$$ is the mass, and $$v$$ is the velocity.

Let the mass of the particle be $$m_p$$ and its velocity $$v_p = 4v_e$$, where $$v_e$$ is the electron's velocity. The de Broglie wavelengths are $$\lambda_p = \frac{h}{m_p v_p}$$ and $$\lambda_e = \frac{h}{m_e v_e}$$.

We are given $$\frac{\lambda_p}{\lambda_e} = \frac{2}{1}$$, so $$\frac{h/(m_p v_p)}{h/(m_e v_e)} = 2$$, which gives $$\frac{m_e v_e}{m_p v_p} = 2$$. Substituting $$v_p = 4v_e$$: $$\frac{m_e v_e}{m_p \times 4v_e} = 2$$, so $$\frac{m_e}{4m_p} = 2$$, giving $$m_p = \frac{m_e}{8}$$.

Therefore the mass of the particle is $$\frac{1}{8}$$ times the mass of the electron.