1.

Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is 5.7 xx 10^5 m//sec (h = 6.626 xx 10^(-34) kg m^2 s^(-1) , mass ofthe electron = 9.1 xx 10^(−31) kg ).

Answer»

Solution :Here we are given
`Delta v = 5.7 xx 10^5 ms^(-1)`
`m = 9.1 xx 10^(-31) kg `
`H = 6.626 xx 10^(-34) kg m^2 s^(-1)`
SUBSTITUTING these values in the equation for uncertainty principle
`Delta x xx (m xx Delta v) = (h)/(4PI)`
we have`Delta x = (h)/(4pi xx m xx Delta v)`
` = (6.626 xx 10^(-34))/(4 xx 22/7 xx 9.1 xx 10^(-31) xx 5.7 xx 10^5)`
` =1.0 xx 10^(-10) m`
Uncertainly in position `= pm 10^(-10) m`


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