AP EAMCET · Chemistry · Structure of Atom
The de' Broglie wavelength of a particle is \(1000 \mathrm{~nm}\). What is its momentum?
\(\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\)
- A \(6.6 \times 10^{-25} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)
- B \(6.6 \times 10^{-25} \mathrm{~J} \mathrm{~s} \mathrm{~m}\)
- C \(6.6 \times 10^{-28} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)
- D \(6.6 \times 10^{-26} \mathrm{~J} \mathrm{~s} \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(6.6 \times 10^{-28} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\lambda=1000 \mathrm{~nm}=1000 \times 10^{-9} \mathrm{~m}=1.0 \times 10^{-6} \mathrm{~m}\) \(\Rightarrow P=\frac{h}{\lambda}=\frac{6.626 \times 10^{-34}}{1.0 \times 10^{-6}}=6.626 \times 10^{-28} \mathrm{~J} \mathrm{~s} \mathrm{~m}^{-1}\)
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