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The de Broglie wavelength associated with a proton under the influence of an electric potential of 100 volts is

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1.227 $A^{0}$

2.86 pm

12.27 $A^{0}$

$1.146 \times 10^{- 21} m$

answer is A.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 m q V}}$
$λ = \frac{6.625 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 1.602 \times 10^{- 19} \times 100}}$

$λ = 1.227 \times 10^{- 10} m = 1.227 \overset{\circ}{A}$

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