If the maximum velocity v of an electron in copper wire is about 2.0x10^8 cm/s, with an uncertainty âĪ 1%, how accurately can the electron's position be determined? In other words, what is the uncertainty in the position of the electron (in meters)?
7 years ago
11 months ago
To determine the uncertainty in the position of the electron, we can use the Heisenberg's uncertainty principle, which states that the uncertainty in the position and momentum of a particle is related by the equation:
ÎðĨ â
Îð âĨ â/4Ï
where ÎðĨ is the uncertainty in position, Îð is the uncertainty in momentum, â is Planck's constant (6.626 x 10^(-34) Jâ
s), and Ï is a mathematical constant approximately equal to 3.14159.
In this case, the uncertainty in momentum can be calculated using the velocity and mass of the electron. The mass of an electron is approximately 9.109 x 10^(-31) kg. The momentum of an electron is given by the equation:
ð = ðâ
ðĢ
where ð is the momentum, ð is the mass, and ðĢ is the velocity.
First, convert the velocity to meters per second:
ðĢ = 2.0 x 10^8 cm/s = 2.0 x 10^6 m/s
Now, calculate the uncertainty in the momentum:
Îð = (1% of ð) = 0.01ð = 0.01â
ðâ
ðĢ
Next, calculate the uncertainty in position:
ÎðĨ â
Îð âĨ â/4Ï
ÎðĨ â
0.01â
ðâ
ðĢ âĨ â/4Ï
ÎðĨ âĨ â/(4Ïâ
0.01â
ðâ
ðĢ)
Finally, plug in the values and calculate the uncertainty in position:
ÎðĨ âĨ (6.626 x 10^(-34) Jâ
s)/(4Ïâ
0.01â
9.109 x 10^(-31) kgâ
2.0 x 10^6 m/s)
ÎðĨ âĨ 1.445 x 10^(-5) meters
Therefore, the uncertainty in the position of the electron is approximately 1.445 x 10^(-5) meters.