Chemistry › Structure of Atom › Dual Behaviour of Matter and Heisenberg Uncertainty Principle
The wavelength of a microscopic particle of mass \(9.1 \times 10^{-31} \mathrm{~kg}\) is \(182…
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 \(\left(h=6.625 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\)
\(728 \times 10^{-23}\) \(7.28 \times 10^{-24}\) \(3.64 \times 10^{23}\) \(3.64 \times 10^{24}\) Solution
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,
\(\begin{aligned}
\lambda & =\frac{h}{m v} \Rightarrow v=\frac{h}{m \times \lambda} \\
v & =\frac{6.625 \times 10^{-34} \mathrm{Js}}{91 \times 10^{-31} \mathrm{~kg} \times 182 \times 10^{-9} \mathrm{~m}} \\
& =0.004 \times 10^6 \mathrm{~m} / \mathrm{s}=4 \times 10^3 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
Therefore, kinetic energy,
\(\begin{aligned}
\mathrm{KE}=\frac{1}{2} m v^2 & =\frac{1}{2} \times 9 \mathrm{I} \times 10^{-31} \times\left(4 \times 10^3\right)^2 \\
& =7.28 \times 10^{-24} \mathrm{~J}
\end{aligned}\)
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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