TS EAMCET · Chemistry · Structure of Atom
The wavelength of an electron is \(10^3 \mathrm{~nm}\). What is its momentum in \(\mathrm{kg} \mathrm{ms}^{-1}\) ? \(\left(h=6.625 \times 10^{-34} \mathrm{~J}\right.\) J s)
- A \(6.625 \times 10^{-31}\)
- B \(6.625 \times 10^{-37}\)
- C \(6.625 \times 10^{-28}\)
- D \(6.625 \times 10^{-34}\)
Answer & Solution
Correct Answer
(C) \(6.625 \times 10^{-28}\)
Step-by-step Solution
Detailed explanation
From de-Broglie wavelength \((\lambda)=\frac{h}{p}\) Give, \(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\) \(\lambda=10^3 \mathrm{~nm}=10^3 \times 10^{-9} \mathrm{~m}=10^{-6} \mathrm{~m}\)…
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