What is the de-Broglie wavelength of the alph | vedclass

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(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.For an $\alpha

Vedclass · Accepted Answer

$\frac{0.101}{\sqrt{V}} \ \mathring{A}$

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(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.For an $\alpha$-particle,the charge $q = 2e = 2 	imes 1.6 	imes 10^{-19} \ C$ and mass $m = 4m_p = 4 	imes 1.67 	imes 10^{-27} \ kg$.Substituting these values:$\lambda = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 4 	imes 1.67 	imes 10^{-27} 	imes 2 	imes 1.6 	imes 10^{-19} 	imes V}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{42.752 	imes 10^{-46} 	imes V}} = \frac{6.63 	imes 10^{-34}}{6.538 	imes 10^{-23} \sqrt{V}} \ m$$\lambda \approx \frac{1.014 	imes 10^{-11}}{\sqrt{V}} \ m = \frac{0.1014 	imes 10^{-10}}{\sqrt{V}} \ m = \frac{0.101}{\sqrt{V}} \ \mathring{A}$.

What is the de-Broglie wavelength of the $\alpha$-particle accelerated through a potential difference $V$?

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