Question

An electron has a momentum py = 2.10 × 10 − 25 kg.m/s. What is the minimum uncertainty in its position that will keep the relative uncertainty in its momentum (Δpy/p) below 2.5%?

Answer 1

Solution)

We mknow,

Delta x* delta P= h/2= h/4pi

Delta Py/Py= 2.5/100

Delta Py=0.025*Py=0.025*(2.10*10^-25 kg.m/s)

=5.25*10^-27 kg-m/s

Now,

The uncertainty in position is,

Delta x= (h/4pi)*1/delta P

=(6.624*10^-34)/4*3.14*(1/5.25*10^-27)

=1.0045*10^-8 m

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