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For a given ion with a charge of $+1$ being accelerated by $U$ volts the distance it travels in a time $t$ is $d$ given by: $$d = v_0 t + \frac{1}{2} a t^2$$ for $v_0=0$ $$a = 2\frac{d}{t^2}$$ and for $a=0$ $$v = \frac{d}{t}$$ therefore $$F = m a = m (2\frac{d}{t^2}) = 2 m \frac{d}{t^2}$$ For the energy of the ion $E_i$ $$E_i = F_i d = (2 m_i \frac{d}{t^2}) d = 2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$ Since $E_v = (1 q_e) U$ where $q_e$ is the universal charge constant it follow that: $$E_i = E_v\ \ =>\ \ \ 2 m_i v_i^2 = q_e U\ \ =>\ \ \ v_i^2 = q_e \frac{U}{2m_i}$$ Therefore: $$\biggl[\ \ \ v_i = \sqrt{ \frac{1}{2} \frac{q_e U}{m_i}}\ \ \ \ \Biggl]$$ However for $v_0 = 0$ $$\operatorname{KE}_i = \frac{1}{2} m_i v_i^2 \ \ \ \text{and} \ \ \ \operatorname{KE}_v = q_e U$$ and since $\operatorname{KE}_i = \operatorname{KE}_v$ is given per definition, since we are talking about the same ion in both cases it follows that

$$\frac{1}{2} m_i v_i^2 = q_e U \ \ \ \ => \ \ \ \ v_i^2 = q_e \frac{U}{\frac{1}2 m_i} = 2 q_e \frac{U}{m_i}$$ therefore

$$\biggl[\ \ \ v_i = \sqrt{ 2 \frac{q_e U}{m_i}}\ \ \ \ \Biggl]$$

What is up with the factor of $\ \sqrt{2}\ \ $ and $\ \ \sqrt{\frac{1}{2}}\ \ $ ???

What am I doing wrong?

Thanks!

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  • $\begingroup$ Your first method is incorrect, because you assumed both a zero and nonzero acceleration at the same time. $\endgroup$
    – Triatticus
    Sep 14 at 21:40
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This part is incorrect:

$$E_i = F_i d = (2 m_i \frac{d}{t^2}) d = 2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$

Specifically $$2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$

is wrong because by your own omission $v_i \neq \frac{d}{t}$

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  • $\begingroup$ Therefore $v_i = \sqrt{2 \frac{q_e U}{m_i}}\ \ \ $ Thanks! $\endgroup$ Sep 14 at 22:34
  • $\begingroup$ and $d = v_0 t + \frac{1}{2} a t^2\ \ $ should be $d = \frac{1}{2} t ( 2 v_0 + a t)\ \ $ and $v = \frac{2 v_0 + a t}{2}\ \ $ and if I substitute $a = \frac{d}{t^2}\ \ $ then I get $2v = v\ \ $ as $ v_0\ \ $ approaches $0\ \ $, so clearly wrong. Thanks! $\endgroup$ Sep 14 at 22:56
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The reason is the way you're making the assumptions of a and v0 to be zero. Either of those is useful on its own but when you do both, you're essentially saying that a=0 and v0=0 at the same time. So then d=0 and F=0 too, so it's a trivial situation where the ion isn't moving and nothing is happening. The odd factors don't actually mean much because they're being multiplied by 0.

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  • $\begingroup$ Very helpful. Thanks! $\endgroup$ Sep 14 at 22:04

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