Answer
It might be a hydrogen-2 nucleus (i.e., a deuteron) or a helium-4 nucleus (i.e., an alpha particle).
Work Step by Step
The particle is undeflected in the crossed fields. Its speed is given by the velocity selector equation, v = E/B. (Section 20–11). If the electric field is removed, the particle travels in a circular path. Calculate the mass of the particle using Newton’s second law.
$$qvB=ma=m\frac{v^2}{r}$$
$$m=\frac{qBr}{v}=\frac{qBr}{(E/B)} $$
We know that the charge q must be an integer multiple of the elementary charge, q = (N)(e).
$$m=\frac{NeB^2r}{E}$$
$$m=\frac{N(1.6\times10^{-19}C)(0.034T)^2(0.027m)}{1500V/m}$$
$$=N(3.3\times10^{-27}kg)\approx N(2.0u)$$
The particle has an atomic mass that is a multiple of 2.0 u.
It might be a hydrogen-2 nucleus (i.e., a deuteron) or a helium-4 nucleus (i.e., an alpha particle).
