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Assuming an electron is confined to a 1n...

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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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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