The de-Broglie wavelength associated with a particle of mass $10^{-6}$ kg with a velocity of $10 \, ms^{-1}$ is:
A$6.63 \times 10^{-22} \, m$
B$6.63 \times 10^{-29} \, m$
C$6.63 \times 10^{-31} \, m$
D$6.63 \times 10^{-34} \, m$
Answer & Solution
Correct answer: B. $6.63 \times 10^{-29} \, m$
Using de Broglie relation, $\lambda=\dfrac{h}{mv}$. Substituting $h=6.626\times10^{-34}\,\text{J s}$, $m=10^{-6}\,\text{kg}$, and $v=10\,\text{m s}^{-1}$ gives $\lambda=\dfrac{6.626\times10^{-34}}{10^{-5}}=6.626\times10^{-29}\,\text{m}$.
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