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A body of mass $10 \mathrm{mg}$ is moving with a velocity of $100 \mathrm{~ms}^{-1}$. The wavelength of de-Broglie wave associated with it would be
( $\left.h=6.63 \times 10^{-34} \mathrm{Js}\right)$
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( $\left.h=6.63 \times 10^{-34} \mathrm{Js}\right)$
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Verified Answer
The correct answer is:
$6.63 \times 10^{-34} \mathrm{~m}$
$m=10 \mathrm{mg}, v=100 \mathrm{~ms}^{-1}, \lambda=$ ?
de-Broglie wavelength, $\lambda=\frac{h}{m v}$
$$
\begin{aligned}
&=\frac{6.63 \times 10^{-34}}{10 \times 10^{-3} \times 100} \\
&=6.63 \times 10^{-34} \mathrm{~m}
\end{aligned}
$$
de-Broglie wavelength, $\lambda=\frac{h}{m v}$
$$
\begin{aligned}
&=\frac{6.63 \times 10^{-34}}{10 \times 10^{-3} \times 100} \\
&=6.63 \times 10^{-34} \mathrm{~m}
\end{aligned}
$$
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