De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to (\(|e|=1.6\times10^{-19}\ \mathrm{C}\), \(m_e=9.1\times10^{-31}\ \mathrm{kg}\), \(h=6.6\times10^{-34}\ \mathrm{J\,s}\))
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- \(0.5\ \mathrm{\AA}\)
- \(1.2\ \mathrm{\AA}\)
- \(1.7\ \mathrm{\AA}\)
- \(2.4\ \mathrm{\AA}\)
The Correct Option is C
Solution and Explanation
\(2meV=2\times9.1\times10^{-31}\times1.6\times10^{-19}\times50=1.456\times10^{-47}\), so \(\sqrt{2meV}=3.816\times10^{-24}\).
\(\lambda=\frac{6.6\times10^{-34}}{3.816\times10^{-24}}=1.73\times10^{-10}\ \mathrm{m}=1.7\ \mathrm{\AA}\).
Answer: (3) 1.7 \(\mathrm{\AA}\)
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