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 = 6.6 xx 10^(-34) kg m^(2) s^(-1)`, mass of the electron `= 9.1 xx 10^(-31) kg`)

Answer» Here, we are given: `Delta v = 5.7 xx 10^(5) m s^(-1), m = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) s^(-1)`
Substiuting these values in the euqation for uncertainty principle, i.e., `Delta x xx (m xx Delta v) = (h)/(4pi)`, we have,
`Delta x = (h)/(4pi xx m xx Delta v) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(4 xx (22)/(7) xx 9.1 xx 10^(-31) kg xx 5.7 xx 10^(5) ms^(-1)) = 1.0 xx 10^(-10) m`
i.e., Uncertainty in position `= +- 10^(-10)m`


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