# Determine KE of electron given momentum & mass [closed]

Some info:

• wavelength of electron: $2.78 \times 10^{-10}$
• momentum of electron: $2.38 \times 10^{-24}$

Determine KE of electron. In a provided hint: $KE = \frac{p^2}{2m}$. So I have:

$$KE = \frac{2.38 \times 10^{-24}}{2 \times 9.11 \times 10^{-31}} = 1.31 \times 10^6$$. But provided answer is $3.10 \times 10^{-18}$. How do I get this?

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## closed as too localized by John Rennie, Waffle's Crazy Peanut, Qmechanic♦Apr 15 '13 at 22:05

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You didn't square the numerator. –  Ataraxia Apr 15 '13 at 6:31
Wow ... that was stupid of me ... –  Jiew Meng Apr 15 '13 at 6:33
lol it happens to the best of us. Well, to me at least :P –  Ataraxia Apr 15 '13 at 6:33
One word for this: UNITS!! Every time I see students do this on assignments I'm marking I risk having a stroke. Without knowing the context (i.e., guessing your wavelength is in meters) I have no way of knowing the relevant physics... i.e. does relativity matter or not, etc.? If meters no, if angstroms yes. :) I'll stop ranting now. –  Michael Brown Apr 15 '13 at 7:13

$p=2.38 \times 10^{-24}\left[\frac{\text {kg m}}{\text s}\right]$
$m_e=9.11\times10^{-31} [\text {kg}]$
\begin{align*} KE = \frac{p^2}{2m} &=\frac{\left(2.38\times10^{-24}\left[\frac{\text{kg m}}{\text s}\right]\right)^2}{2\times9.11\times10^{-31}[\text{kg}]}\\&=\frac{2.38^2\times10^{-24\times2}\left[\frac{\text{kg m}}{\text s}\right]^2}{2\times9.11\times10^{-31}[\text{kg}]}\\ &=\frac{5.6644\times10^{-48}}{18.22\times10^{-31}}\left[\frac{\text{kg m}^2}{\text s^2}\right]\\ &=\frac{5.6644}{18.22}\times10^{-48+31}[\text J]\\ &=0.310889\times10^{-17}[\text J]\\&=3.11\times10^{-18}[\text J] \end{align*}.