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An electron is accelerated through a potential difference of 100 V. Calculate the de Broglie wavelength associated with it. (Take h=6.63\times 10^{-34}\, \mathrm{Js},\, m_e=9.1\times 10^{-31}\, \mathrm{kg},\, e=1.6\times 10^{-19}\, \mathrm{C}).
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Given:
- Potential difference, V = 100 V
- Planck's constant, h = 6.63 × 10-34 Js
- Mass of electron, me = 9.1 × 10-31 kg
- Charge of electron, e = 1.6 × 10-19 C
The de Broglie wavelength (λ) associated with an electron accelerated through a potential difference V is given by:
λ = h√2meeV
Substituting the values:
λ = 6.63 × 10-34√2 × 9.1 × 10-31 × 1.6 × 10-19 × 100
λ = 6.63 × 10-34√29.12 × 10-48
λ = 6.63 × 10-345.396 × 10-24
λ ≈ 1.227 × 10-10 m
Converting to Angstroms (1 Å = 10-10 m):
λ ≈ 1.227 Å
(Alternatively, using the shortcut formula λ ≈ 12.27 / √V Å: λ = 12.27 / 10 = 1.227 Å)
Therefore, the de Broglie wavelength associated with the electron is approximately 1.227 Å (or 0.123 nm).
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