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What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

Jayant Kumar , 10 Years ago
Grade 11
anser 1 Answers
Apoorva Arora

Last Activity: 10 Years ago

KE= 120 eV
P=\sqrt{2mKE}
=5.9\times 10^{-24}
P=mv
so v=6.5\times 10^{6}m/s
\lambda =\frac{h}{P}
= 1.123 angstrom
Thanks and Regards
Apoorva Arora
IIT Roorkee
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