Answer Speed of Electron in Hydrogen's Ground State

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SUMMARY

The discussion centers on calculating the speed of an electron in the ground state of hydrogen, given the uncertainty in its position and velocity. The relevant equation used is derived from the Heisenberg Uncertainty Principle: h/(4πmΔv). The constants provided are Planck's constant (h = 6.63e-34), the electron mass (m = 9.11e-31), and the uncertainty in position (Δx = 0.10 nm). The correct approach reveals that the speed of the electron is approximately 5.8e5 m/s, correcting the misconception that the speed equals the uncertainty in position.

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Homework Statement



In the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.10 nm. If the speed of the electron is approximately the same as the uncertainty in its speed, about how fast is it moving?

Homework Equations


h/(4pi*m*v) =
h=6.63e-34
m=9.11e-31
v=.10e-9




The Attempt at a Solution



I thought you should just substitute in everything in the equation...

this is what I'm sure the answer is..but it was marked wrong..

6.63e-34/(4pi*9.11e-31*(.10e-9) = 5.8e5= 1e6..however it is marked wrong..please help!
 
Physics news on Phys.org
velocity does not equal 0.1e-9... you're trying to FIND the velocity. It is just saying that the uncertainty in the velocity is approximately the velocity, i.e.
<br /> \Delta v \approx v<br />
 

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