What is the potential difference needed to accelerate a He+ ion (charge +e; 4 amu) from rest to a speed of 2*10^6 m/s?

Question

anonymous · Answer

Okay, so this should just be an energy problem right? Since it starts from rest$$K=U$$ and $$K=\frac{1}{2}mv^{2}$$and $$U=qV$$ substitute solve for $$V=\frac{mv^{2}}{2q}$$ Correct?

prakharluv · Answer

Let potential difference = V
q V = (1/2)*m*(vf^2 - vi^2)
Can you substitute values and calculate V?

anonymous · Answer

Yes. That is what I did, except when I substituted the values in I had:
$$q = +1.602*10^{-19}C$$$$m = 4*1.67*10^{-27} kg= 6.68*10^{-27}kg$$$$v_{f}-v_{i}=0m/s-2*10^{6}m/s=-2*10^6m/s$$Plug those values in I have $$V=\frac{mv^2}{2q}=\frac{(6.68*10^{-27}kg)*(-2*10^6m/s)^{2}}{2*(1.602*10^{-19}C)}=83395.7 V$$...oh my gosh this is embarrassing. I was forgetting to square the delta v term whenever I plugged it in last night. So I kept getting .041 V .. lol. Thanks.