The uncertainty in a proton\'s position is 0.19nm. What is the minimum uncertain

Question

The uncertainty in a proton's position is 0.19nm. What is the minimum uncertainty delta p in its momentum? delta p min = _____ kg m/s What is the kinetic energy of a proton whose momentum is equal to this uncertainty (delta p=p)? K=_____meV

Explanation / Answer

uncertainty in a proton's position*uncertainty in a proton's momentum = h/4*pi =>uncertainty in a proton's momentum = (6.626*(10^(-34)))/(4*pi()*0.19*(10^(-9))) = 2.7752*(10^(-25)) kg m/s = p We know that K.E. = 0.5*m*v^2 = (0.5*m^2*v^2)/m = p^2/2m =>K.E. = ((2.7752*(10^(-25)))^2)/(2*1.67*(10^(-27))) = 2.31E-23 J = (2.31E-23/(1.60217646*(10^(-13)))) = 1.44*(10^(-10)) MeV

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.