The de-Broglie wavelength of an electron havi | vedclass

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(D) Kinetic energy,$KE = 80 eV = 80 	imes 1.6 	imes 10^{-19} J = 128 	imes 10^{-19} J$.De-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{

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$1.4$

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(D) Kinetic energy,$KE = 80 eV = 80 	imes 1.6 	imes 10^{-19} J = 128 	imes 10^{-19} J$.De-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.Substituting the values:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 128 	imes 10^{-19}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{256 	imes 9 	imes 10^{-50}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{16 	imes 3 	imes 10^{-25}}$$\lambda = \frac{6.6}{48} 	imes 10^{-9} m = 0.1375 	imes 10^{-9} m = 1.375 	imes 10^{-10} m$.Since $1 	ext{ Å} = 10^{-10} m$,we get $\lambda \approx 1.4 	ext{ Å}$.

The de-Broglie wavelength of an electron having $80 eV$ energy is nearly ($1 eV = 1.6 \times 10^{-19} J$,Mass of the electron $= 9 \times 10^{-31} kg$,Planck's constant $= 6.6 \times 10^{-34} J-s$) (in $text{ Å}$)

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