1.

For an electron v= 300 ms^(-1) andcertaintyin velocityis 0.001 % whatis theuncertainty in position

Answer»

`5.76xx 10^(2) m`
`1.92 xx 10^(2) m`
`3.84 xx 1^(2) m`
`19.2 xx 10^(-2) m`

Solution :`Delta Xxxm Delta v = (H)/(4PI)`
`Delta x= (h)/(4pim Delta v )`
`Delta v = 0.001 %of300 `
`=(300 xx0.001)/(100)`
`m=9.1 xx 10^(31) kg`
`=(6.626 xx 10^(34))/(4xx 3.14 xx 9.1xx 10^(31)xx 300 xx 0.001 xx 10^(2))`
`=1.932 xx 10^(2) m`


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