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(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.Since kinetic energy $K = \frac{p^2}{2m}$, we have $p = \sqrt{2mK}$.Thus,

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$124.2$

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(B) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$.Since kinetic energy $K = \frac{p^2}{2m}$, we have $p = \sqrt{2mK}$.Thus, $\lambda = \frac{h}{\sqrt{2mK}}$.Given $h = 4.14 	imes 10^{-15} eVs$, $K = 100 eV$, and $m = \frac{0.5 	imes 10^6}{c^2} eV/c^2$.Substituting these values, where $c = 3 	imes 10^8 m/s$:$\lambda = \frac{4.14 	imes 10^{-15}}{\sqrt{2 	imes (0.5 	imes 10^6 / c^2) 	imes 100}}$$\lambda = \frac{4.14 	imes 10^{-15} 	imes c}{\sqrt{10^8}}$$\lambda = \frac{4.14 	imes 10^{-15} 	imes 3 	imes 10^8}{10^4} = 1.242 	imes 10^{-10} m = 124.2 pm$.

The de-Broglie wavelength of an electron having kinetic energy $100 eV$ is, $[$ Use $h=4.14 \times 10^{-15} eVs$, mass of electron $= \frac{0.5 \times 10^6}{c^2} eV/c^2$, $1 pm = 10^{-12} m$ $]$ (in $pm$)

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