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Q: 11.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be $1.40\times 10^-^1^0\hspace{1mm}m$ ?

Answers (1)

S Sayak

For the given wavelength momentum of the neutron will be p given by

$\\p=\frac{h}{\lambda }\\ p=\frac{6.62\times 10^{-34}}{1.4\times 10^{-10}}\\ p=4.728\times 10^{-24}kg\ m\ s^{-1}$

The kinetic energy K would therefore be

$\\K=\frac{p^{2}}{2m}\\ K=\frac{(4.728\times 10^{-24})^{2}}{2\times 1.675\times 10^{-27}}\\ K=6.67\times 10^{-21}J$

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