Chemistry - Atomic Structure Question with Solution | TestHub

The de-Broglie wavelength associated with the charged particle as

$\text{λ=}\frac{\text{h}}{\text{mv}}$

$\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=}\text{K}\text{.E}$

$\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{×}\text{2m}\text{=}\text{2m}\left(\text{K}\text{.E}\right)\text{=}{\text{m}}^{\text{2}}{\text{v}}^{\text{2}}$

$\sqrt{\text{2m·}\left(\text{K}\text{.E}\right)}\text{=}\text{mv}$

$\lambda =\frac{\text{h}}{\sqrt{2\text{m}\left(\text{K.E}\right)}}$

$=\frac{\text{h}}{\sqrt{2{\text{m}}_{\alpha }·2eV}}$

$=\frac{\text{6.626}\times 10^{-34}}{\sqrt{2\times \text{1.66}\times 10^{-27}\times 4\times \text{1.6}\times 10^{-19}\times 2\times \text{V}}}$

$=\frac{\text{6.626}\times 10^{-34}}{\sqrt{\text{V}}\times 10^{-23}\times \text{6.519}}$

$=\frac{\text{1.018}\times 10^{-11}}{\sqrt{\text{V}}}$

$\text{=}\frac{\text{0}\text{.1018}}{\sqrt{\text{V}}}\stackrel{\text{o}}{\text{A}}$