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KCET · Physics · Ray Optics

Find the de-Broglie wavelength of an electron with kinetic energy of \( 120 \mathrm{eV} \).

  1. A \( 95 \mathrm{pm} \)
  2. B \( 102 \) pm
  3. C \( 112 \mathrm{pm} \)
  4. D \( 124 \mathrm{pm} \)
Verified Solution

Answer & Solution

Correct Answer

(C) \( 112 \mathrm{pm} \)

Step-by-step Solution

Detailed explanation

Given, kinetic energy of electron \(=120 \mathrm{eV}\)
de Broglie wavelength is related to \(\mathrm{V}\) as
\(\lambda=\frac{1.227 \mathrm{~nm}}{\sqrt{V}}=\frac{1.227 \times 10^{-9}}{\sqrt{120}}=\frac{1.227 \times 10^{-9}}{10.95}\)
\(=0.112 \times 10^{-9}\)
Therefore,
\(\lambda=112 \times 10^{-12} \mathrm{~m}=112 \mathrm{pm}\)