The kinetic energy of a proton wave is 500 e | vedclass

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant $(6.626 	imes 10^{-34

Vedclass · Accepted Answer

$4.953 	imes 10^{-12} \ m$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant $(6.626 	imes 10^{-34} \ J \cdot s)$,$m$ is the mass of the proton $(1.67 	imes 10^{-27} \ kg)$,and $K$ is the kinetic energy in Joules.First,convert kinetic energy from $eV$ to Joules: $K = 500 \ eV 	imes 1.602 	imes 10^{-19} \ J/eV = 8.01 	imes 10^{-17} \ J$.Now,substitute the values into the formula: $\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 1.67 	imes 10^{-27} 	imes 8.01 	imes 10^{-17}}}$.$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2.675 	imes 10^{-43}}} = \frac{6.626 	imes 10^{-34}}{1.635 	imes 10^{-22}} \approx 4.05 	imes 10^{-12} \ m$.

The kinetic energy of a proton wave is $500 \ eV$. Calculate its de Broglie wavelength. (Mass of proton $= 1.67 \times 10^{-27} \ kg$)

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