Question:
The de-Broglie wavelength of a 10 KeV electron is equal to:
The de-Broglie wavelength of a 10 KeV electron is equal to:
Updated On: Mar 26, 2025
- \( 12.3 \text{Å} \)
- \( 1.23 \text{Å} \)
- \( 0.123 \text{Å} \)
- \( 123 \text{Å} \)
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The Correct Option is B
Solution and Explanation
The de-Broglie wavelength is given by:
\[
\lambda = \frac{h}{\sqrt{2mE}}
\]
For a 10 KeV electron, using standard values for \( h \) and \( m \), we get:
\[
\lambda \approx 1.23 \text{Å}
\]
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