Question

An alpha particle (2 protons + 2 neutrons) is brought to a stop by a potential difference of 450 V . What was its initial speed?


Express your answer using two significant figures.

Answer 1

The work done to stop the particle is $A = q V$. Since the mechanical energy is conserved, $T_i - A = T_f$, where $T_{i,f}$ is the kinetic energy of alpha particle in the initial and final state: $T_i = \frac{m v_i^2}{2}$ , $T_f = 0$.

The charge of the alpha particle is $q = 2 e = 3.2 \cdot 10^{-19} C$ , and the mass of the alpha particle is $6.64 \cdot 10^{-27} kg$.

Hence, the initial speed was $v_i = \sqrt{\frac{2 A}{m}} = \sqrt{\frac{2 q V}{m}} = 208.26 \cdot 10^6 \frac{m}{s}$.