Question

The uncertainty in a proton's position is 0.015 nm. a) What is the minimum uncertainty Δp in its momentum? b) What is the kinetic energy of a proton whose momentum is equal to this uncertainty (Δp=p)? (Unit: meV)

Answer 1

Answer:

$\Delta P_{x}=3.51 *10^{-24}$ kg m/s

$E = 3.68 * 10^{-21}$ J

Explanation:

given data

uncertainty in proton position $\Delta x$ = 0.015 nm

according to Heisenberg's principle of uncertainty

$\Delta x \Delta P_{x}= \frac{\frac{h}{2\pi }}{2}$

Where h is plank constant = $6.6260 * 10^{-34}$ j-s

$\Delta P_{x}= \frac{\frac{h}{2\pi }}{2\Delta x}$

$\Delta P_{x}= \frac{\frac{6.6260 * 10^{-34}}{2\pi }}{2*0.015*10^{-9}}$

$\Delta P_{x}=3.51 *10^{-24}$ kg m/s

b) kinetic energy  of proton whose momentum

$P =\Delta p$

$E =\frac{\Delta p^{2}}{2m}$

where m is mass is proton

$E =\frac{(3.51*10^{-24})^{2}}{2*1.67*10^{-27}}$

$E = 3.68 * 10^{-21}$ J