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An electron having velocity 2xx10^(6)m//s has uncertainty in kineticenergy is (6.66)/pixx10^(-21) J, then calculate the uncretainty in position (in Angstrom ,Å) of the electron .[Given :h=6.60xx10^(-34) J-sec] |
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Answer» SOLUTION :`KE=(1)/(2) mv^(2)""V=2xx10^(6)` `d(KE)=mvdv` `dv=(d(KE))/(mv)""......(1)""But""DELTA"x"=(h)/(4pimDeltav)""....(2)` `Delta"x"=(h)/(4pim(d(KE))/(mv))"" , ""Delta"x"=(6.62xx10^(-34)xx2xx10^(6))/(4pixx6.62/(pi)xx10^(-21))=500Å` |
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