The de Broglie wavelength associated with a b | vedclass

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Question

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}

Vedclass · Accepted Answer

$6.626 	imes 10^{-34} \, m$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the de Broglie equation: $\lambda = \frac{h}{\sqrt{2mE}}$.Given $m = 1 \, kg$,$E = 0.5 \, J$,and $h = 6.626 	imes 10^{-34} \, J \cdot s$:$\lambda = \frac{6.626 	imes 10^{-34}}{\sqrt{2 	imes 1 	imes 0.5}} = \frac{6.626 	imes 10^{-34}}{\sqrt{1}} = 6.626 	imes 10^{-34} \, m$.

The de Broglie wavelength associated with a ball of mass $1 \, kg$ having kinetic energy $0.5 \, J$ is

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