KCET · Physics · Dual Nature of Matter
The de Broglie wavelength of an electron accelerated to a potential of \( 400 \mathrm{~V} \) is approximately
- A \( 0.03 \mathrm{~nm} \)
- B \( 0.04 \mathrm{~nm} \)
- C \( 0.12 \mathrm{~nm} \)
- D \( 0.06 \mathrm{~nm} \)
Answer & Solution
Correct Answer
(D) \( 0.06 \mathrm{~nm} \)
Step-by-step Solution
Detailed explanation
Given, electron is accelerated to a potential of \( 400 \mathrm{~V} \), then de Broglie wavelength is related to potential as
\[
\begin{array}{l}
\lambda=\frac{1.227 \mathrm{~nm}}{\sqrt{V}}=\frac{1.227 \mathrm{~nm}}{\sqrt{400}}=\frac{1.227}{20} \mathrm{~nm} \\
\Rightarrow \lambda=0.06 \mathrm{~nm}
\end{array}
\]
Therefore, de Broglie wavelength is \( 0.06 \mathrm{~nm} \).
\[
\begin{array}{l}
\lambda=\frac{1.227 \mathrm{~nm}}{\sqrt{V}}=\frac{1.227 \mathrm{~nm}}{\sqrt{400}}=\frac{1.227}{20} \mathrm{~nm} \\
\Rightarrow \lambda=0.06 \mathrm{~nm}
\end{array}
\]
Therefore, de Broglie wavelength is \( 0.06 \mathrm{~nm} \).
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