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JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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The momentum of a particle which has a de-Broglie wavelength of

0.1 \mathrm{~nm}

is

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Correct option: D

6.62 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}

Option D (

6.62 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}

) is correct because it is calculated directly from the de-Broglie wavelength formula.

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