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For an electron v= 300 ms^(-1) andcertaintyin velocityis 0.001 % whatis theuncertainty in position |
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Answer» `5.76xx 10^(2) m` `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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