The de-Broglie wavelength of an electron having 80 eV of energy is nearly leV -1-6 - 10-19 J- Mass of electron 9 - 10-31 kg and Plank-s constant 6-6 - 10-34 J sec A- 0-14 AB- 14 AC- 140 AD- 1-4 A

The correct option is D 1.4∘ABy using λ =h/√(2mE)=12.27/√(V). If energy is 80 eV then accelerating potential difference will be 80 V. So λ =12.27/√(80)=1.37≈ ...

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Standard XII

Physics

Matter Waves

The de-Brogli...

Question

The de-Broglie wavelength of an electron having 80eV of energy is nearly $(l e V = 1.6 \times 10^{- 19} J$ Mass of electron $9 \times 10^{- 31} k g$ and Plank's constant $6.6 \times 10^{- 34} J s e c)$

140 \circ A

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0.14 \circ A

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14 \circ A

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1.4 \circ A

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Solution

The correct option is D $1.4 \circ A$
By using $λ = \frac{h}{\sqrt{2 m E}} = \frac{12.27}{\sqrt{V}}$ . If energy is 80 eV then accelerating potential difference will be 80 V. So
$λ = \frac{12.27}{\sqrt{80}} = 1.37 \approx 1.4 \circ A .$

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Q. The de-Broglie wavelength of an electron having 80eV of energy is nearly

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

The correct option is D 1.4∘ABy using λ=h√2mE=12.27√V. If energy is 80 eV then accelerating potential difference will be 80 V. So λ=12.27√80=1.37≈1.4∘A.

The de-Broglie wavelength of an electron having 80eV of energy is nearly $(l e V = 1.6 \times 10^{- 19} J$ Mass of electron $9 \times 10^{- 31} k g$ and Plank's constant $6.6 \times 10^{- 34} J s e c)$

The correct option is D $1.4 \circ A$
By using $λ = \frac{h}{\sqrt{2 m E}} = \frac{12.27}{\sqrt{V}}$ . If energy is 80 eV then accelerating potential difference will be 80 V. So
$λ = \frac{12.27}{\sqrt{80}} = 1.37 \approx 1.4 \circ A .$