An atom of a particular element is traveling at $1 \%$ of the speed of light. The de Broglie wavelength is found to be $3.31 \times 10^{-3} \mathrm{pm} .$ Which element is this?
Added by Alberto T.
Step 1
Step 1: Use the de Broglie wavelength formula $\lambda = \frac{h}{p}$, where $\lambda$ is the de Broglie wavelength, $h$ is the Planck constant, and $p$ is the momentum of the particle. Show more…
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