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Assuming an electron is confined to a 1nm wide region,find the uncertainty in momentum using Heisenberg Uncertainty principle .You can assume the uncertainty in position Deltax as 1 nm.Assuming p~~Deltap, find th energy of the electron in electron volts. |
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Answer» SOLUTION :According to Heisenberg.s uncertainty principle, `(Deltax)(Deltap)~~H` `therefore (Deltax)(Deltap)~~(h)/(2PI)` `therefore Deltap~~(h)/(2piDeltax)` `therefore Deltap=(6.625xx10^(-34))/(2xx3.14xx1xx10^(-9))` `therefore Deltap=1.055xx10^(-25)NS` From the statement ,we have `Deltap`=p and so p=`1.055xx10^(-25)Ns` Energy of GIVEN electron, `E=(p^(2))/(2m)` `therefore =((1.055xx10^(-25))^(2))/(2xx9.1xx10^(-31))` `therefore E=(1.113xx10^(-50))/(18.2xx10^(-31))` `therefore E=6.115xx10^(-21)J` `therefore E=(6.115xx10^(-21))/(1.6xx10^(-19))eV` `therefore E=3.822xx10^(-2)eV` |
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