Question

An alpha particle (a bare helium nucleus with (q=+2e) accelerates across a 100 V potential difference, starting from rest. What is the particle’s kinetic energy in eV when it reaches the negative electrode? This question requires no mathematics beyond what you can do in your head.

Short answer

100ev

Step-by-step solution

given information

In this given

the value of charge is Q = 2q

and the potential diff is $\nabla$V = 100ev

and we have to find the kinetic energy of particle ??

simplify

it is a very simple formula based question . the value of q and v have given put in the formula of

K.E. = $Q\times$$\nabla$V

solution

now, we have to put the value of q and v in the formula of K.E.

K.E. = $Q\times$$\nabla$V

$$\begin{gathered} Q=2q \\ \nabla V\hspace{0.17em}=100ev \\ K.E.=2\times 100 \\ K.E.=200ev \end{gathered}$$

hence the kinetic energy of particle is 200ev .