The de-Broglie wavelength of an electron having 80 eV of energy is nearly(1 ev =1.6×10-19 J, Mass of electron =9×10-31kg Plank's constant=6.6×10-34 J-sec)

λ=h2mE =6.6×10-342×9×10-31×80×1.6×10-19=1.4Ao

The de-Broglie wavelength of an electron having 80 eV of energy is nearly(1 ev =1.6×10-19 J, Mass of electron =9×10-31kg Plank's constant=6.6×10-34 J-sec)

λ=h2mE =6.6×10-342×9×10-31×80×1.6×10-19=1.4Ao

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

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$140 \overset{o}{A}$

$0.14 \overset{o}{A}$

$14 \overset{o}{A}$

$1.4 \overset{o}{A}$

answer is D.

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Detailed Solution

$λ = \frac{h}{\sqrt{2 mE}} = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}} = 1.4 \overset{o}{A}$

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The de-Broglie wavelength of an electron having 80 eV of energy is nearly(1 ev =1.6×10-19 J, Mass of electron =9×10-31kg Plank's constant=6.6×10-34 J-sec)

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