Question:

What will be the kinetic energy of an electron having de-Broglie wavelength 2Å?

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$E = \frach²2mλ²}$ for electron kinetic energy.
Updated On: Apr 16, 2026
  • 37.5 eV
  • 75 eV
  • 150 eV
  • 300 eV
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The Correct Option is A

Solution and Explanation


Step 1:
$E = \frach²2mλ² = \frac(6.63×10^-34)²2 × 9.1×10^-31 × (2×10^-10)² = 6.0×10^-18$ J.

Step 2:
Convert to eV: $\frac6.0×10^-181.6×10^-19 = 37.5$ eV.
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