AP EAMCET · Chemistry · Structure of Atom
The wavelength of a microscopic particle of mass \(9.1 \times 10^{-31} \mathrm{~kg}\) is \(182 \mathrm{~nm}\), its kinetic energy in \(\mathrm{J}\) is \(\left(h=6.625 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\)
- A \(728 \times 10^{-23}\)
- B \(7.28 \times 10^{-24}\)
- C \(3.64 \times 10^{23}\)
- D \(3.64 \times 10^{24}\)
Answer & Solution
Correct Answer
(B) \(7.28 \times 10^{-24}\)
Step-by-step Solution
Detailed explanation
Given, mass of particles \(=9.1 \times 10^{-31} \mathrm{~kg}\) Wavelength \((\lambda)=182 \mathrm{~nm}=182 \times 10^{-9} \mathrm{~m}\) According to de-Broglie equation,…
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