Q132 / 180 · NEET 2020 (Phase 2) · Physics

The de-Broglie wavelength of an electron moving with kinetic energy of $144 eV$ is nearly

Correct answer
A
$102 \times 10^{- 3} nm$
B
$102 \times 10^{- 4} nm$
C
$102 \times 10^{- 5} nm$
D
$102 \times 10^{- 2} nm$

Solution

The formula of de-Broglie wavelength, $λ = \frac{h}{\sqrt{2 m K . E .}}$ and we know that, $K . E . = e V$ . Recall the proper relation between wavelength and electric potential, in angstrom, $λ = \frac{12.27}{\sqrt{V}} Å$ $= \frac{12.27}{\sqrt{144}} \times 10^{- 10}$ $= 1.02 \times 10^{- 10} m$

it means the wavelength, $λ = 102 \times 10^{- 3} nm$ .

Hence, the correct answer is option: 'A'.

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The de-Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly

Solution

Relevant Questions

The de-Broglie wavelength of an electron moving with kinetic energy of $144 eV$ is nearly