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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$.Given $KE = 2.5 \ eV = 2.5 	imes 1.6 	imes 10^{-19}

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$\frac{h 	imes 10^{25}}{\sqrt{72}}$

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$.Given $KE = 2.5 \ eV = 2.5 	imes 1.6 	imes 10^{-19} \ J = 4 	imes 10^{-19} \ J$.Given $m_{e} = 9 	imes 10^{-31} \ kg$.Substituting the values:$\lambda = \frac{h}{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 4 	imes 10^{-19}}}$$\lambda = \frac{h}{\sqrt{72 	imes 10^{-50}}}$$\lambda = \frac{h}{\sqrt{72} 	imes 10^{-25}}$$\lambda = \frac{h 	imes 10^{25}}{\sqrt{72}}$.

The de Broglie wavelength of an electron with kinetic energy of $2.5 \ eV$ is (in $m$):
$(1 \ eV = 1.6 \times 10^{-19} \ J, m_{e} = 9 \times 10^{-31} \ kg)$

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The de Broglie wavelength of an electron with kinetic energy of $2.5 \ eV$ is (in $m$):$(1 \ eV = 1.6 \times 10^{-19} \ J, m_{e} = 9 \times 10^{-31} \ kg)$