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(C) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Given:$h = 6.6 	imes 10^{-34}\; Js$$m

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$1.2$

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(C) The de Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Given:$h = 6.6 	imes 10^{-34}\; Js$$m = 9.1 	imes 10^{-31}\; kg$$E = 100\; eV = 100 	imes 1.6 	imes 10^{-19}\; J = 1.6 	imes 10^{-17}\; J$.Substituting these values into the formula:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 1.6 	imes 10^{-17}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{29.12 	imes 10^{-48}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{5.396 	imes 10^{-24}}$$\lambda \approx 1.22 	imes 10^{-10}\; m$.Since $1\; \mathring{A} = 10^{-10}\; m$,the wavelength is approximately $1.2\; \mathring{A}$.

An electron beam has a kinetic energy equal to $100\; eV$. Find its wavelength (in $\mathring{A}$) associated with the beam,given the mass of an electron $= 9.1 \times 10^{-31}\; kg$,$1\; eV = 1.6 \times 10^{-19}\; J$,and Planck's constant $= 6.6 \times 10^{-34}\; Js$.

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