Question

assume an electron in the ground state of the hydrogen atom moves at an average speed of 5.00 × 10^6 m/s. if the speed is known to an uncertainty of 1%, what is the minimum uncertainty in its position? round your answer to 3 significant digits. the radius of the hydrogen atom in the ground state is 5.29 × 10^-11 m. m

Explanation:

Step1: Calculate the uncertainty in velocity

The uncertainty in velocity \(\Delta v\) is \(1\%\) of the average speed \(v\).
\(\Delta v = 0.01\times5.00\times 10^{6}\ \text{m/s}=5.00\times 10^{4}\ \text{m/s}\)

Step2: Use the Heisenberg uncertainty principle

The Heisenberg uncertainty principle is \(\Delta x\Delta p\geq\frac{h}{4\pi}\), and since \(p = mv\) (for non - relativistic case, \(m = 9.11\times 10^{-31}\ \text{kg}\) for an electron), \(\Delta p=m\Delta v\). Then \(\Delta x\geq\frac{h}{4\pi m\Delta v}\), where \(h = 6.626\times 10^{-34}\ \text{J}\cdot\text{s}\)
Substitute the values:

$$
LATEXBLOCK0
$$

$$
LATEXBLOCK1
$$

Answer:

\(1.16\times 10^{-9}\ \text{m}\)