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Question: The de-Broglie wavelength of an electron having 80*eV* of energy is nearly (\(1eV = 1.6 \times 10^{-...

The de-Broglie wavelength of an electron having 80eV of energy is nearly (1eV=1.6×10−19J1eV = 1.6 \times 10^{- 19}J, Mass of electron 9×10−31kg9 \times 10^{- 31}kg and Plank's constant 6.6×10−346.6 \times 10^{- 34}J-sec)

A

140 Å

B

0.14 Å

C

14 Å

D

1.4 Å

Answer

1.4 Å

Explanation

Solution

By using λ=h2mE=12.27V\lambda = \frac{h}{\sqrt{2mE}} = \frac{12.27}{\sqrt{V}}.

If energy is 80 eV then accelerating potential difference will be 80 V. So λ=12.2780=1.37≈1.4A˚.\lambda = \frac{12.27}{\sqrt{80}} = 1.37 \approx 1.4Å.