Question

the momentum of an electron is measured as 4.80*10^-27 kg m /s with an uncertainty of 2.4*10^-29 kg m /s. what is the minimum uncertainty in the determination of its position? (h = 6.63*10^-34 J s? options 3.20 * 10*-5, 4.40 * 10^-6

Answer 1

As per the principle of Heisenberg uncertainty

we know that

$\Delta x * \DeltaP = \frac{h}{2\pi}$

now here we know that

$\Delta P$ = uncertainty in momentum = $2.4 * 10^{-29} kg*m/s$

$\Delta x$ = uncertainty in position

now using the above equation we have

$2.4 * 10^{-29} * \Delta x = \frac{6.63 * 10^{-34}}{2\pi}$

$\Delta x = \frac{6.63 * 10^{-34}}{2.4 * 10^{-29}* 2\pi}$

$\Delta x = 4.4 * 10^{-6} m$

so the uncertainty in position will be given by above value.