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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 |
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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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