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  4. Debroglie wavelength of protons accelerated by an electric field at a potential difference V is

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Q. Debroglie wavelength of protons accelerated by an electric field at a potential difference V is

Dual Nature of Radiation and Matter

Solution:

$k \cdot E=q V=e V=e V$
$P=\sqrt{2 m k \cdot E}=\sqrt{2 m e v}$
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m e v}}$
$\frac{6.62 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6818^{19} \times V}}$
$\lambda=\frac{6.62 \times 10^{-34}}{\sqrt{3.2 \times 1.67} \times 10^{-23}}$
$=\frac{6.62}{\sqrt{3.2 \times 1.67}} \times 10^{-10} \times 10^{-1}$
$=\frac{0.662}{\sqrt{3.2 \times 1.67}} A^{\circ}$
$\lambda=\frac{0.662}{2.31 \sqrt{v}} A^{\circ}$
$=\frac{0.286}{\sqrt{v}} A^{\circ}$