1
$\begingroup$

How can I express speed in Angstroms/second after application of a force that's in eV/Angstrom (given the mass, original velocity and time elapsed)?

Improve this question
$\endgroup$
14
  • 1
    $\begingroup$ This is impossible, because the force and mass only determines the acceleration (via F=ma), but not the speed. Besides, "velocity" is the correct word here; "speed" may be an acceptable word but it is at least not professional $\endgroup$
    – wzkchem5
    May 29, 2022 at 11:35
  • $\begingroup$ @wzkchem5 and speed/velocity is a * t, right? So if I also know the time it should be possible? I am just very confused in those small measurements. $\endgroup$
    – John T
    May 29, 2022 at 14:38
  • 1
    $\begingroup$ No. The change of velocity within an infinitesimal time dt is a*dt. The relation does not hold if the time interval is not infinitesimal. So you have to know (1) the original velocity, and (2) the time period; moreover, the time period has to be extremely short. With those in hand, you can calculate the velocity right after that extremely short time period. $\endgroup$
    – wzkchem5
    May 29, 2022 at 14:48
  • 1
    $\begingroup$ @Sha The poster gets to choose which site to post their question. On MMSE, we only migrate questions to other SE sites if the question is struggling to find an answer here (or is likely to struggle to get an answer here). Likewise, you could say that many questions on Physics.SE "belong" here at MMSE (just see for example, the questions in their DFT tag, and the fact that 30% of all MMSE questions have the DFT tag, so it's "our territory"), but that's just how this complex network of Q/A sites works :) $\endgroup$
    – Nike Dattani
    May 29, 2022 at 18:44
  • 1
    $\begingroup$ This is basically a homework question. Do we answer those here? $\endgroup$
    – B. Kelly
    May 29, 2022 at 19:09

1 Answer 1

Reset to default
1
$\begingroup$

Applying a force $F$ acting on a mass $m$, the acceleration according to classical mechanics is:

$$\tag{1} a = \frac{F}{m}. $$

Given the speed $v(0)$ at time $t=0$ and the acceleration, we can obtain the speed at any time by one of the "big four" kinematic equations which are usually taught in high school physics, and most certainly taught in every first-year university physics curriculum on the planet. The relevant kinematic equation for your question is:

$$\tag{2} v(t) = v(0) + a\cdot t. $$

Substituting Eq. (1) into Eq. (2) gives us:

$$\tag{3} v(t) = v(0) + \frac{F}{m}t. $$

If $v(0)$ is given to you in Angstroms/second, we simply need to ensure that $Ft/m$ is also in Angstroms/second. If $F$ is given to you in eV/Angstrom, $m$ is given to you in kg, and $t$ is being measured in seconds, we can start by converting the Force into SI units:

$$\tag{4}{\tiny F ~~\textrm{[eV/Angstrom}] \times \left( \frac{1.60218\times10^{-19} \textrm{J}}{\textrm{eV}}\right) \times \left( 1 ~~\frac{\textrm{kg}\cdot {\textrm{m}}^2/\textrm{s}^2}{{\textrm{J}}}\right)\times \left(\frac{10^{10} \textrm{Angstroms}}{\textrm{m}}\right)^2 = F~~\textrm{[kg $\cdot$ Angstroms/s$^2$]}}.$$

So your force in eV/Angstrom just needs to be multiplied by 16.0218 to get a force in kg$\cdot$Angstroms/s$^2$. You can then multiply it by the time in seconds, and divide by the mass in kg, and you will get the "change in speed" in Angstroms/s from the starting speed of $v(0)$ which is also given in Angstroms/s, therefore yielding a final speed in Angstroms/s.

Note that all of this assumes that the acceleration is constant, which means the force needs to be constant.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .