quantum numbers

Question

An electron is accelerated through a potential difference of 150 V. Its de Broglie wavelength is closest to (in nm): (h = 6.626×10⁻³⁴ J s, m_e = 9.11×10⁻³¹ kg, 1 eV = 1.602×10⁻¹⁹ J)

RankGuru · Accepted Answer

1. **Wavelength Formula**: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m E_k}}$.
2. **Accelerated Electron**: Energy $E_k = qV$, where $q$ is electronic charge.
3. **Shortcut for Electron**: $\lambda (	ext{in Å}) \approx \sqrt{\frac{150}{V}}$.
4. **Apply Shortcut**:
   - $V = 150$ V.
   - $\lambda \approx \sqrt{\frac{150}{150}} = 1.0$ Å.
5. **Unit Conversion**: $1.0$ Å $= 10^{-10}$ m $= 0.1$ nm.
6. **Result**: The wavelength is **$0.10$ nm**.

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