Question

The de-Broglie wavelength \( \lambda \) associated with an electron of energy \( V \) electron volt is:

• A. \( \frac{1.227}{\sqrt{V}} \) nm
• B. \( \frac{0.1227}{\sqrt{V}} \) nm
• C. \( 1.227 \) nm
• D. \( \frac{12.27}{\sqrt{V}} \) nm

Hint:

Higher kinetic energy reduces the de-Broglie wavelength, following \( \lambda \propto 1/\sqrt{V} \).

Answer 1

The Correct Option is A

The de-Broglie wavelength formula for an electron with kinetic energy \( V \) eV is:
\[ \lambda = \frac{h}{\sqrt{2m e V}} \] For an electron, this simplifies to:
\[ \lambda = \frac{1.227}{\sqrt{V}} \text{ nm}. \]