1. 12.2×10⁻¹³ m 2. 12.2×10⁻¹² m 3. 12.2×10⁻¹⁴ m 4. 12.2 nm

Answer: 2. 12.2×10⁻¹² m de Broglie wavelength of an electron accelerated through V : λ=h/√(2meV) . Substituting h=6.63×10⁻³⁴ , m=9×10⁻³¹ , e=1.6×10⁻¹⁹ , V=10⁴ : sqrt2×9×10⁻³¹×1.6×10⁻¹⁹×10⁴=sqrt2.88×10⁻⁴⁵≈5.4×10⁻²³ . λ≈dfrac6.63×10⁻³⁴5.4×10⁻²³≈1.22×10⁻¹¹ m =12.2×10⁻¹² m.

Topic: Matter waves, de Broglie relation· Dual Nature of Matter and Radiation

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (m_e = 9×10^-31 kg)

1. \(12.2\times10^{-13}\) m

2. \(12.2\times10^{-12}\) m

3. \(12.2\times10^{-14}\) m

4. 12.2 nm

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Answer: 2. \(12.2\times10^{-12}\) m

de Broglie wavelength of an electron accelerated through \(V\): \(\lambda=\dfrac{h}{\sqrt{2meV}}\).

Substituting \(h=6.63\times10^{-34}\), \(m=9\times10^{-31}\), \(e=1.6\times10^{-19}\), \(V=10^4\):

\(\sqrt{2\times9\times10^{-31}\times1.6\times10^{-19}\times10^4}=\sqrt{2.88\times10^{-45}}\approx5.4\times10^{-23}\).

\(\lambda\approx\dfrac{6.63\times10^{-34}}{5.4\times10^{-23}}\approx1.22\times10^{-11}\) m \(=12.2\times10^{-12}\) m.

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An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : (m_e = 9×10^-31 kg)

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