Question

German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δ????) Δx≥ℎ4????mΔ???? where ℎ is Planck's constant and m is the mass of the object. The mass of an electron is 9.11×10−31 kg. What is the uncertainty in the position of an electron moving at 4.00×106 m/s with an uncertainty of Δ????=0.01×106 m/s?

Answer 1

`\Delta x = 5.47 * 10^(-9) m`

Step-by-step explanation:

As we know by the principle of uncertainty that the product of uncertainty in position and uncertainty in momentum is given as

`\Delta x * \Delta P = (h)/(4\pi)`

so here we know that

`\Delta v = 0.01 * 10^6 m/s`

`m = 9.11 * 10^(-31) kg`

so we have

`\Delta x * (9.11 * 10^(-31))(0.01 * 10^6) = (6.26 * 10^(-34))/(4\pi)`

`\Delta x = 5.47 * 10^(-9) m`