What is the de Broglie wavelength (in mathrin | 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 acquired by th

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

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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 acquired by the particle.For an $\alpha$-particle,the charge $q = 2e$ and mass $m = 4m_p \approx 4 	imes 1.67 	imes 10^{-27} 	ext{ kg}$.Substituting these values into the formula: $\lambda = \frac{h}{\sqrt{2(4m_p)(2eV)}} = \frac{h}{\sqrt{16m_p eV}} = \frac{h}{4\sqrt{m_p eV}}$.Using the constants $h = 6.626 	imes 10^{-34} 	ext{ J s}$,$m_p = 1.67 	imes 10^{-27} 	ext{ kg}$,and $e = 1.6 	imes 10^{-19} 	ext{ C}$:$\lambda = \frac{6.626 	imes 10^{-34}}{4 \sqrt{1.67 	imes 10^{-27} 	imes 1.6 	imes 10^{-19} 	imes V}} 	ext{ m}$.Converting to $\mathring{A}$ $(1 	ext{ m} = 10^{10} \mathring{A})$:$\lambda \approx \frac{0.101}{\sqrt{V}} \mathring{A}$.

What is the de Broglie wavelength (in $\mathring{A}$) of an $\alpha$-particle accelerated through a potential difference of $V$ volts?

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