Question
For the de Broglie wavelength of 10−17 m , momentum of a particle will be
For the de Broglie wavelength of 10−17 m , momentum of a particle will be
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Solution
The correct option is C 
\{p} =\frac{h}{\lambda}=(\frac{6.62\times 10^{-34}}{10^{-17}})=6.6 \times 10^{-17}\)
\{p} =\frac{h}{\lambda}=(\frac{6.62\times 10^{-34}}{10^{-17}})=6.6 \times 10^{-17}\)
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