de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to ( |e| = 1.6 × 10 - 19 C, me = 9.1 × 10 - 31 kg, h = 6.6 × 10 - 34 Js): (a) 0.5 (b) 1.2 (c) 1.7 (d) 2.4

Answer: (c). λ = (h)/(√(2meV)). √2×9.1×10⁻³¹×1.6×10⁻¹⁹×50 = √1.456×10⁻⁴⁷ ≈ 3.82×10⁻²⁴ kg m/s. λ = 6.6×10⁻³⁴3.82×10⁻²⁴ ≈ 1.7×10⁻¹⁰ m. So the answer is (c) 1.7 Å.

de-Broglie wavelength of an electron accelerated by a voltage of 50 V… | JEE Main 2015 (10 Apr, Shift 1)

Q135JEE Main 2015 · 10 APR · SHIFT 1Easy

Matter Waves: de Broglie Wavelength

de-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to ( $|e|$ = 1.6 $\times$ 10 $-$ 19 C, me = 9.1 $\times$ 10 $-$ 31 kg, h = 6.6 $\times$ 10 $-$ 34 Js):

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Answer: (c)

$\lambda = \dfrac{h}{\sqrt{2meV}}$ .
$\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times50} = \sqrt{1.456\times10^{-47}} \approx 3.82\times10^{-24}$ kg m/s.
$\lambda = \dfrac{6.6\times10^{-34}}{3.82\times10^{-24}} \approx 1.7\times10^{-10}$ m.
So the answer is (c) 1.7 Å.

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