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If a proton is accelerated through a potential difference of $1000 \mathrm{~V}$, then its de-Broglie wavelength is (given, $m_p=1.67 \times 10^{-27} \mathrm{~kg}$, $\left.h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
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The correct answer is:
$9.1 \times 10^{13} \mathrm{~m}$
$\lambda_{\text {proton }}=\frac{h}{p}=\frac{0.286}{\sqrt{V}}(Å)$
(This formula is for protons.)
$$
\therefore \quad \lambda=\frac{0.286}{\sqrt{1000}} \times 10^{-10} \mathrm{~m} \approx 9.1 \times 10^{-13} \mathrm{~m}
$$
(This formula is for protons.)
$$
\therefore \quad \lambda=\frac{0.286}{\sqrt{1000}} \times 10^{-10} \mathrm{~m} \approx 9.1 \times 10^{-13} \mathrm{~m}
$$
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