Question

# A proton in a linear accelerator has a de Broglie wavelength of 122 pm. What is the speed of the proton?

Solution

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Using the de Broglie equation, which is: $\lambda= \dfrac{h}{mv}$ Where:

$\lambda$= wavelength

h= Planck's constant which is 6.626$\times10^{-34}$ J $\cdot$ s

m= mass of the object

v= velocity or the speed of the object

Given the wavelength which is 122 pm or 122 $\times10^{-12}$ meters sine pico(p) is equal to $10^{-12}$. The mass of a single proton is 1.67$\times10^{-27}$ kg. Thus finding the velocity our equation should be, $v= \dfrac{h}{m\lambda}$ $v= \dfrac{6.626\times10^{-34}\text{ J $\cdot$ s}}{(1.67\times10^{-27} \text{ kg})(122\times10^{-12}\text{ m})}$ $v=\boxed{ 3252.18\text{ m/s}}$

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