1.

An electron and a photon each have a wavelength of 1.00nm. Find (i) their momentum (ii) the energy of the photon and (iii) K.E of electron

Answer»

Solution :`lambda_(e)=lambda_("photon")=1.00 nm=10^(-9)m`
(i) For electron or photon, momentum
`p=p_(e)=p_(p)=(h)/(lambda)`
`p=((6.63xx10^(-34)))/(10^(-9))=6.63xx10^(-25)kg m//s`
(ii) Energy of photon
`E=(HC)/(lambda)`
`=((6.63xx10^(-34))xx(3xx10^8))/(10^(-9))J`
`~~19.89xxxx10^(-17)J`
`(~~1243eV)`
(iii) Kinetic energy of electron `=(p^(2))/(2M)`
`(1)/(2)xx((6.63xx10^(-25)))/(9.1xx10^(-31))J ~~ 2.42xx10^(-19)J`
`(~~1.51eV)`


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