If a CRT needs both location and momentum of electrons to faithfully render an image on its screen how does The Uncertainty Principle (TUP) reconcile this. It seems TUP is a "catch-all" for Quantum Physics (QP) to ensure 100% predictability ["QP has neer been wrong"].

Let's run the numbers and see if the Heisenburg Uncertainty Principle is inconsistent with how a CRT works:

The equation is:

\sigma _x \sigma _p \ge \frac {\hbar} 2

where the sigmas are the uncertainty in position and momentun, respectively. Given the mass of an electron and typical velocity from an electron gun of about 6x10^6 m/s, the momentum of the electron is:

p = 9.11 \times 10^{-31} Kg \ \times \ 6 \times 10^6 \frac m s = 5.5 \times 10^{-24} \frac {Kg-m} s

The variability of velocity is probably around plus or minus 1%, given the variation in how electrons boil off a cathode and how accurately the voltage of the elctron gun is regulated. But for fun let's assume that the variability is held to a very high standard, say 0.01%. This means that the uncertainty of the electron's momentum is:

\sigma_p = 0.0001\ \times \ 5.5 \times 10^{-24} \frac {Kg-m} s = 5.5 \times 10^{-28} \frac {Kg-m} s

So from Heisenburg the uncertainty of the electron's position is:

\sigm_x\ \ge \ \frac {\hbar}{2 \sigma_p} \ = \ \frac {1.04 \times 10^{-34} J-s}{2 (5.5 \times 10^{-28} \frac {Kg-m} s ) } = 9.6 \times 10^{-8} m

This level of uncertainty in position of an electron is much, much smaller than the width of the electron beam hitting the CRT screen, so the Heisenburg Uncertainty Principle does not adversely affect the sharpness of the CRT display.

Let's run the numbers and see if the Heisenburg Uncertainty Principle is inconsistent with how a CRT works:

The equation is:

\sigma _x \sigma _p \ge \frac {\hbar} 2

where the sigmas are the uncertainty in position and momentun, respectively. Given the mass of an electron and typical velocity from an electron gun of about 6x10^6 m/s, the momentum of the electron is:

p = 9.11 \times 10^{-31} Kg \ \times \ 6 \times 10^6 \frac m s = 5.5 \times 10^{-24} \frac {Kg-m} s

The variability of velocity is probably around plus or minus 1%, given the variation in how electrons boil off a cathode and how accurately the voltage of the elctron gun is regulated. But for fun let's assume that the variability is held to a very high standard, say 0.01%. This means that the uncertainty of the electron's momentum is:

\sigma_p = 0.0001\ \times \ 5.5 \times 10^{-24} \frac {Kg-m} s = 5.5 \times 10^{-28} \frac {Kg-m} s

So from Heisenburg the uncertainty of the electron's position is:

\sigm_x\ \ge \ \frac {\hbar}{2 \sigma_p} \ = \ \frac {1.04 \times 10^{-34} J-s}{2 (5.5 \times 10^{-28} \frac {Kg-m} s ) } = 9.6 \times 10^{-8} m

This level of uncertainty in position of an electron is much, much smaller than the width of the electron beam hitting the CRT screen, so the Heisenburg Uncertainty Principle does not adversely affect the sharpness of the CRT display.
Then you mean the probability of the velocity is in a specific tolerance? How is the electron impeaded or accelerated to meet the screen at a specific place with the required intensity by either electrically charged plates and/or a magnetic field.

Sorry nykkyo, I'm not following your question. Are you asking how does an electron gun work to accelerate electrons to a certain velocity? Or are you asking how does a CRT combine the electron gun with electrostatic or magnetic deflection to aim the beam at a specific spot on the CRT screen?

Sorry nykkyo, I'm not following your question. Are you asking how does an electron gun work to accelerate electrons to a certain velocity? Or are you asking how does a CRT combine the electron gun with electrostatic or magnetic deflection to aim the beam at a specific spot on the CRT screen?
Just unsuccessfully sayig statistics can prove anything!