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Q: 11.15 (a) What is the de Broglie wavelength of a bullet of mass 0.040\hspace{1mm}kg travelling at the speed of  1.0\hspace{1mm}km/s

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The momentum of the bullet is

\\p=mv\\ p=0.04\times 10^{3}\\ p=40\ kg\ m\ s^{-1}

De Broglie wavelength is

\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{40}\\ \lambda =1.655\times 10^{-35}\ m

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