MHT CET · Chemistry · Structure of Atom
What is momentum of a microscopic particle having de Broglie's wavelength \(6.0 Å\) ? ( \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) )
- A \(4.6 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- B \(1.1 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- C \(3.18 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}\)
- D \(6.36 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(1.1 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
According to the de Broglie's equation,
\(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)
\(\therefore \mathrm{p} =\frac{\mathrm{h}}{\lambda}=\frac{6.63 \times 10^{-34} \mathrm{Js}}{6.0 \times 10^{-10} \mathrm{~m}}=1.1 \times 10^{-24}\)\(\mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
\(\lambda=\frac{\mathrm{h}}{\mathrm{p}}\)
\(\therefore \mathrm{p} =\frac{\mathrm{h}}{\lambda}=\frac{6.63 \times 10^{-34} \mathrm{Js}}{6.0 \times 10^{-10} \mathrm{~m}}=1.1 \times 10^{-24}\)\(\mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}\)
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