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A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\left.\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
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(Plank's Constant $\left.\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
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Verified Answer
The correct answer is:
$1 \times 10^{-21} \mathrm{~J}$
As $\lambda=\frac{\mathrm{h}}{\mathrm{P}} \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{~K} . \mathrm{E})}} \Rightarrow \mathrm{K} \cdot \mathrm{E}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^2}$
So, K.E $=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 2 \times 10^{-27} \times\left(3.3 \times 10^{-10}\right)^2}$
$=1 \times 10^{-21} \mathrm{~J}$
So, K.E $=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 2 \times 10^{-27} \times\left(3.3 \times 10^{-10}\right)^2}$
$=1 \times 10^{-21} \mathrm{~J}$
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