An alpha particle (mass 6.6 x 10^-24 grams) emitted by radium travels at 3.4 x 10^7 plus or minus 0.1 x 10^7 mi/h.
(a) What is its de Broglie wavelength (in meters)?
(b) Whats uncertainty in its position?
for part (a) I got 1.53 meters which said it was wrong on Aris and I got 1.56 x 10^-56 meters for part (b) the uncertainty which said it was wrong too.
This is how I solved it please tell me what I did wrong:
part (a) lambda =h/mu
first convert mi/hr to meters/sec
3.4 x 10^7mi/hr = 1.52 x 10^7m/s
(6.626 x 10^-34 kg m^2 s)/(6.6 X 10^-27 kg)x(1.52 x 10^7 m/s)
= 1.53 meters
part (b) delta(x) x m delta u is greater than or equal to h/4 pi.
so, delta (x) is greater than or equal to (h)/ (4 pi) x (m) delta (u)
delta x= (6.626 x 10^-34 kg m^2 s)/ (4 pi) x (6.6 x 10^-27 kg)(4.48 x 10^5 m/s)
= 1.56 x 10^-55 meters
I also converted the mass 6.6 x 10^-24 grams into kilograms, making it 6.6 x 10^-27 kilograms.
Please look this over and correct my error, thank you.
I think you have converted from miles/hour to m/s correctly. But when I put that into the equation, which you appear to have set up correctly, I don't get your answer but something like 6 x 10^-15 or so meters.
Run that calculation through your calculator again and see what you get.
Your mistake lies in the conversion of mass from grams to kilograms. You have correctly converted 6.6 x 10^-24 grams to 6.6 x 10^-27 kilograms, but you incorrectly used 6.6 x 10^-27 kilograms instead of 6.6 x 10^-27 kg in your calculations for both parts (a) and (b).
Here's the corrected solution:
(a) To find the de Broglie wavelength of the alpha particle, you can use the equation:
λ = h / (mu)
where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 kg m^2/s), m is the mass of the particle (6.6 x 10^-27 kg), and u is the velocity of the particle (1.52 x 10^7 m/s).
Substituting the values into the equation:
λ = (6.626 x 10^-34 kg m^2/s) / ((6.6 x 10^-27 kg) * (1.52 x 10^7 m/s))
= 6.626 x 10^-34 / (6.6 x 10^-27 * 1.52 x 10^7) m
= 1.53 x 10^-15 m (rounded to 3 significant figures)
Therefore, the de Broglie wavelength of the alpha particle is approximately 1.53 x 10^-15 meters.
(b) To find the uncertainty in the position of the alpha particle, you can use the uncertainty principle equation:
Δx Δp ≥ h / (4π)
where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant (6.626 x 10^-34 kg m^2/s).
Since the uncertainty in momentum (Δp) is given by the mass (m) times the uncertainty in velocity (Δu), we have:
Δp = m Δu = (6.6 x 10^-27 kg) * (4.48 x 10^5 m/s) = 2.9568 x 10^-21 kg m/s (rounded)
Substituting the values into the uncertainty principle equation:
Δx * (2.9568 x 10^-21 kg m/s) ≥ (6.626 x 10^-34 kg m^2/s) / (4π)
Δx ≥ (6.626 x 10^-34 kg m^2/s) / ((4π) * (2.9568 x 10^-21 kg m/s))
≈ 1.42 x 10^-14 m (rounded to 3 significant figures)
Therefore, the uncertainty in the position of the alpha particle is approximately 1.42 x 10^-14 meters.