I asked the professor .7759 is correct
Hello!
This problem is similar to the previous. The uncertainty principle states that both position and velocity of a particle cannot be measured exactly. Mathematically,
Delta p*Delta x gt= bar h/2,
where p is for impulse (momentum), which is equal to m*v, mass by velocity, and x is the position. Correspondingly, Delta p means uncertainty of measuring momentum and Delta x means uncertainty of measuring position. Obviously Delta p = m*Delta v.
bar h is the so-called reduced Planck's constant, h/(2 pi). Therefore the minimum uncertainty in the speed is
Delta v = h/(4pi)*1/(m*Delta x).
All quantities are given, so the numerical result is
Delta v approx (6.626*10^(-34))/(4*3.14)*1/(6.10*10^(-25)*0.7*10^(-9)) =(6.626)/(4*3.14*6.10*0.7) approx0.1235 (m/s).
I took into account that nano- means 10^(-9).
To obtain the result you want one should "forget" to divide by 2pi, then it would be about 0.7759 m/s, C. But I'm almost sure about the values.
The second question, about the minimum speed, requires a separate consideration.
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