The de-Broglie wavelength of an electron havi | vedclass

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(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.Given values are:$h = 6.6 	imes 10^{-34} \ J \cdot

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$1.4$

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(D) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mE}}$.Given values are:$h = 6.6 	imes 10^{-34} \ J \cdot s$$m = 9 	imes 10^{-31} \ kg$$E = 80 \ eV = 80 	imes 1.6 	imes 10^{-19} \ J = 128 	imes 10^{-19} \ J = 1.28 	imes 10^{-17} \ J$.Substituting these values into the formula:$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9 	imes 10^{-31} 	imes 1.28 	imes 10^{-17}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{\sqrt{23.04 	imes 10^{-48}}}$$\lambda = \frac{6.6 	imes 10^{-34}}{4.8 	imes 10^{-24}}$$\lambda = 1.375 	imes 10^{-10} \ m \approx 1.375 \ \mathring{A} \approx 1.4 \ \mathring{A}$.Thus,the correct option is $D$.

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

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