Question

If the position of an electron in an atom is measured to an accuracy of 0.0100 nm, what is the
electron's uncertainty in velocity?​

Answer 1

Answer:

the position of electron is measured to an occurrence of 0.0100 and the electronic honestly in the vertically are 20.10

Answer 2

Answer:

The uncertainty in the velocity of the electron, ΔV, calculated is $5.79 \times 10^{6}m/s$.

Explanation:

Data given,

The uncertainty in the position of an electron, Δx = $0.0100nm$ = $0.01 \times 10^{-9} m$= $10^{-11}m$.

The uncertainty in the velocity of the electron, ΔV =?

As we know, according to the uncertainty principle:

• $\Delta x \times \Delta p = \frac{h}{4\pi }$

Here, Δp = uncertainty in momentum = $m\Delta V$

Here, m = mass of electron = $9.1\times 10^{-31}kg$

• h = Planck's constant = $6.626\times 10^{-34} kg-m^{2}/s$

Now, the equation becomes:

• $\Delta x \times m\Delta V = \frac{h}{4\pi }$
• $\Delta V = \frac{h}{4\pi m \Delta x}$

After substituting the values in the equation, we get:

• $\Delta V = \frac{6.626\times10^{-34} }{4\times 3.14\times 9.1\times 10^{-31} \times 10^{-11} }$
• $\Delta V = \frac{6.626\times10^{8} }{4\times 3.14\times 9.1} }$
• $\Delta V = \frac{6.626\times10^{8} }{114.296 }$
• $\Delta V = 0.0579\times10^{8}$ = $5.79 \times 10^{6}$ m/s

Hence, the uncertainty in the velocity of the electron, ΔV, calculated is $5.79 \times 10^{6}m/s$.