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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} \, J \cdot s$$m = 9.1 	imes

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$6.068 	imes 10^{-9} \, m$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} \, J \cdot s$$m = 9.1 	imes 10^{-31} \, kg$$v = 1.2 	imes 10^5 \, ms^{-1}$Substituting these values:$\lambda = \frac{6.626 	imes 10^{-34}}{(9.1 	imes 10^{-31}) 	imes (1.2 	imes 10^5)}$$\lambda = \frac{6.626 	imes 10^{-34}}{10.92 	imes 10^{-26}}$$\lambda \approx 0.6068 	imes 10^{-8} \, m = 6.068 	imes 10^{-9} \, m$.

The de Broglie wavelength of an electron moving with a velocity of $1.2 \times 10^5 \, ms^{-1}$ is ...... .

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