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If an electron has wavelength of 1Å, find the momentum?
1. \(6.62 \times {10^{ - 24}}\;kg.m.{s^{ - 1}}\)
2. \(6.62 \times {10^{ - 30}}\;kg.m.{s^{ - 1}}\)
3. \(6.62 \times {10^{24}}\;kg.m,{s^{ - 1}}\)
4. None

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Correct Answer - Option 1 : \(6.62 \times {10^{ - 24}}\;kg.m.{s^{ - 1}}\)

For an electron of momentum p has de-Broglie wavelength,

\(\begin{array}{l} \lambda = \frac{h}{p}\\ \Rightarrow p = \frac{h}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}}}}{{{{10}^{ - 10}}}} = 6.62 \times {10^{ - 24}}\;kg.m.{s^{ - 1}} \end{array}\)

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