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: λ=hmv\lambda= \dfrac{h}{mv} Where:

λ\lambda= wavelength

h= Planck's constant which is 6.626×1034\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 ×1012\times10^{-12} meters sine pico(p) is equal to 101210^{-12}. The mass of a single proton is 1.67×1027\times10^{-27} kg. Thus finding the velocity our equation should be, v=hmλv= \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=3252.18 m/sv=\boxed{ 3252.18\text{ m/s}}

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