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AP EAMCET · Chemistry · Structure of Atom

The de' Broglie wavelength of a particle is \(1000 \mathrm{~nm}\). What is its momentum?
\(\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\)

  1. A \(6.6 \times 10^{-25} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)
  2. B \(6.6 \times 10^{-25} \mathrm{~J} \mathrm{~s} \mathrm{~m}\)
  3. C \(6.6 \times 10^{-28} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)
  4. D \(6.6 \times 10^{-26} \mathrm{~J} \mathrm{~s} \mathrm{~m}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(6.6 \times 10^{-28} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\lambda=1000 \mathrm{~nm}=1000 \times 10^{-9} \mathrm{~m}=1.0 \times 10^{-6} \mathrm{~m}\) \(\Rightarrow P=\frac{h}{\lambda}=\frac{6.626 \times 10^{-34}}{1.0 \times 10^{-6}}=6.626 \times 10^{-28} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)