An -particle is accelerated from rest through a potential difference of 6.0V. Its de-Broglie wavelen — Atomic Structure Chemistry Question

💡 Solution & Explanation

$\lambda = \frac{h}{\sqrt{2mqV}} = \frac{6.626 \times 10^{-34}}{\sqrt{2(4 \times 1.66 \times 10^{-27})(2 \times 1.6 \times 10^{-19}) \times 6}} = \frac{6.626 \times 10^{-34}}{\sqrt{2.55 \times 10^{-44} \times 6}} = \frac{6.626 \times 10^{-34}}{\sqrt{15.3 \times 10^{-44}}} \approx \frac{6.626 \times 10^{-34}}{3.91 \times 10^{-22}} \approx 1.69 \times 10^{-12}$ m. Wait, formula simplifies to $\lambda = \frac{0.101}{\sqrt{V}}$ \AA for alpha. $\lambda = \frac{0.101}{\sqrt{6}} \text{ \AA} = 0.0412 \text{ \AA} = 4.12$ pm. Therefore, correct answer is B.