Question:
The momentum of an electron is \(6.625 \times 10^{-28} \, kg \cdot ms^{-1}\). What is its wavelength in nm?
- (Given: \( h = 6.625 \times 10^{-34} \, J \cdot s\))
The momentum of an electron is \(6.625 \times 10^{-28} \, kg \cdot ms^{-1}\). What is its wavelength in nm?
- (Given: \( h = 6.625 \times 10^{-34} \, J \cdot s\))
Show Hint
The de Broglie wavelength of a particle is given by:
\[
\lambda = \frac{h}{p}
\]
Always ensure units are consistent, and convert meters to nanometers by multiplying by \(10^9\).
Updated On: May 5, 2026
- \(100\)
- \(1000\)
- \(500\)
- \(6625\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: de Broglie wavelength formula
\[ \lambda = \frac{h}{p} \]
\[ h = 6.625 \times 10^{-34} \, \text{J·s}, \quad p = 6.625 \times 10^{-28} \, \text{kg·m s}^{-1} \]
Step 2: Substitute values
\[ \lambda = \frac{6.625 \times 10^{-34}}{6.625 \times 10^{-28}} = 10^{-6} \, \text{m} \]
Step 3: Convert to nm
\[ 1 \, \text{m} = 10^9 \, \text{nm} \]
\[ \lambda = 10^{-6} \times 10^9 = 10^3 \, \text{nm} = 1000 \, \text{nm} \]
Final Answer: \( 1000 \, \text{nm} \)
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