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Question

(C) The de Broglie wavelength $\lambda$ associated with a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given

Vedclass · Accepted Answer

$\lambda _d = \frac{0.202}{\sqrt{V}} \ \mathring{A}$

Vedclass · Answer

(C) The de Broglie wavelength $\lambda$ associated with a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.For a deuteron, the mass $m_d \approx 2m_p$ (where $m_p$ is the mass of a proton) and the charge $q_d = e$.Substituting the values: $m_d = 2 	imes 1.67 	imes 10^{-27} \ 	ext{kg} = 3.34 	imes 10^{-27} \ 	ext{kg}$ and $q_d = 1.6 	imes 10^{-19} \ 	ext{C}$.Using the formula $\lambda = \frac{h}{\sqrt{2m_d e V}}$, we get $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 3.34 	imes 10^{-27} 	imes 1.6 	imes 10^{-19} 	imes V}}$.Calculating the constant: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{10.688 	imes 10^{-46} 	imes V}} = \frac{6.63 	imes 10^{-34}}{3.27 	imes 10^{-23} \sqrt{V}} \approx \frac{2.02 	imes 10^{-11}}{\sqrt{V}} \ 	ext{m}$.Converting to $\mathring{A}$ $(1 \ \mathring{A} = 10^{-10} \ 	ext{m})$: $\lambda = \frac{0.202}{\sqrt{V}} \ \mathring{A}$.

The de Broglie wavelength for a deuteron can be given by:

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