An α–particle is accelerated from rest through a potential difference of 6.0V. Its de Broglie wavel

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An α–particle is accelerated from rest through a potential di­fference of 6.0V. Its de Broglie wavelength is

Accepted Answer

$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m2V}}$

$= \frac{6.626 \times 10^{- 34}}{\sqrt{2 \times 4 \times 1.66 \times 10^{- 27} \times 2 \times 1.6 \times 10^{- 19} \times 6}} = 4.15 \times 10^{- 12}\text{ m}$

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