The de Broglie wavelength associated with a p | vedclass

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(B) The kinetic energy $K$ gained by a proton of mass $m$ and charge $e$ accelerated through a potential $V$ is given by $K = eV$.Since $K = \frac{p^2

Vedclass · Accepted Answer

$2.86 \ pm$

Vedclass · Answer

(B) The kinetic energy $K$ gained by a proton of mass $m$ and charge $e$ accelerated through a potential $V$ is given by $K = eV$.Since $K = \frac{p^2}{2m}$,we have $p = \sqrt{2meV}$.The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.Substituting the values: $h = 6.626 	imes 10^{-34} \ J \cdot s$,$m = 1.67 	imes 10^{-27} \ kg$,$e = 1.6 	imes 10^{-19} \ C$,and $V = 100 \ V$.$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 1.67 	imes 10^{-27} 	imes 1.6 	imes 10^{-19} 	imes 100}}$$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{5.344 	imes 10^{-44}}}$$\lambda = \frac{6.626 	imes 10^{-34}}{2.3117 	imes 10^{-22}} \approx 2.866 	imes 10^{-12} \ m = 2.86 \ pm$.

The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

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