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)
A
140∘A
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B
0.14∘A
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C
14∘A
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D
1.4∘A
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Solution
The correct option is D1.4∘A By 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.