The de Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly (a) 102 × 10⁻² nm (b) 102 × 10⁻³ nm (c) 102 × 10⁻⁴ nm (d) 102 × 10⁻⁵ nm

Answer: (b). For an electron, λ = (1.227)/(√(V)) nm, where K = eV, so V = 144 V. λ = (1.227)/(12) ≈ 0.102 nm. So the answer is (b) 102×10⁻³ nm.

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Matter Waves: de Broglie Wavelength

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

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Answer: (b)

For an electron, $\lambda = \dfrac{1.227}{\sqrt{V}}$ nm, where $K = eV$ , so $V = 144$ V.
$\lambda = \dfrac{1.227}{12} \approx 0.102$ nm.
So the answer is (b) $102\times10^{-3}$ nm.

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