Answer:
K = 1.29eV
Explanation:
In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:
[tex]\Delta x \Delta p\geq \frac{h}{4\pi}[/tex] (1)
Δx : uncertainty of position = 2.0pm = 2.0*10^-12m
Δp: uncertainty of momentum = ?
h: Planck's constant = 6.626*10^-34 J.s
You calculate the minimum possible value of Δp from the equation (1):
[tex]\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}[/tex]
The minimum kinetic energy is calculated by using the following formula:
[tex]k=\frac{(\Delta p)^2}{2m}[/tex] (2)
m: mass of the proton = 1.67*10^{-27}kg
[tex]k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J[/tex]
in eV you have:
[tex]2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV[/tex]
The kinetic energy of the proton is 1.29eV
The kinetic energy of the proton is mathematically given as
K.E=1.29eV
Question Parameters:
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis.
Generally, the equation for the uncertainty principle is mathematically given as
[tex]dxdp * \d p\geq \frac{h}{4\pi}[/tex]
Where the minimum possible value of Δp
[tex]dp=\frac{h}{4\pi d x}\\\\dp=\frac{6.626*10^{-34}}{4\pi(2.0*10^{-12})}\\\\ dp=2.63*10^{-23}kg.\frac{m}{s}[/tex]
The minimum kinetic energy
[tex]k=\frac{( dp)^2}{2m}[/tex]
[tex]k=\frac{(2.63*10^{-23})^2}{2(1.67*10^{-27})}\\\\k=2.08*10^{-19}J[/tex]
In conclusion, what is the kinetic energy of the proton is
K.E=2.08*10^{-19}*6.242*10^{18}eV
K.E=1.29eV
Read more about Kinetic energy
https://brainly.com/question/999862
