1.

Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Answer»

SOLUTION :`lambda=(H)/(sqrt(2mE))`
`=(6.625 XX 10^(-34))/(sqrt(2 xx 9.1 xx 10^(31) xx 120 xx 1.6 xx 10^(-19)))`
`=1.121 xx 10^(-10)`m.


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