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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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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
up vote 2 down vote accepted

$p=2.38 \times 10^{-24}\left[\frac{\text {kg m}}{\text s}\right]$

$m_e=9.11\times10^{-31} [\text {kg}]$

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

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(+1) Excellent answer. :-) – Parth Kohli Apr 15 '13 at 11:20
While I truly encourage this answer, I'm quite suspicious whether it helps solving it..? :O – Waffle's Crazy Peanut Apr 15 '13 at 12:38
-1 While this answer is extremely helpful to the poster it is in no way helpful to anyone else or the site. – Brandon Enright Apr 15 '13 at 19:51

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