KCET · Physics · Ray Optics
Find the de-Broglie wavelength of an electron with kinetic energy of \( 120 \mathrm{eV} \).
- A \( 95 \mathrm{pm} \)
- B \( 102 \) pm
- C \( 112 \mathrm{pm} \)
- D \( 124 \mathrm{pm} \)
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,
\(\lambda=112 \times 10^{-12} \mathrm{~m}=112 \mathrm{pm}\)
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,
\(\lambda=112 \times 10^{-12} \mathrm{~m}=112 \mathrm{pm}\)
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