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An electron having velocity `2xx10^(6)m//s` has uncertainty in kinetic energy 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]`

Answer» `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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