What will be the de Broglie wavelength associ | vedclass

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Question

(B) The de Broglie wavelength $\lambda$ is given by $\lambda = h / p = h / \sqrt{2mE_k}$.For a charged particle accelerated through a potential differ

Vedclass · Accepted Answer

$0.286 / \sqrt{V}$

Vedclass · Answer

(B) The de Broglie wavelength $\lambda$ is given by $\lambda = h / p = h / \sqrt{2mE_k}$.For a charged particle accelerated through a potential difference $V$,the kinetic energy $E_k = qV$.Thus,$\lambda = h / \sqrt{2mqV}$.For a proton,$m = 1.67 	imes 10^{-27} \ kg$ and $q = 1.602 	imes 10^{-19} \ C$.Substituting these values: $\lambda = 6.626 	imes 10^{-34} / \sqrt{2 	imes 1.67 	imes 10^{-27} 	imes 1.602 	imes 10^{-19} 	imes V}$.$\lambda \approx 0.286 / \sqrt{V} \ \mathring{A}$.

What will be the de Broglie wavelength associated with a charged proton in $\mathring{A}$? ($V =$ potential difference)

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