An alpha particle (mass = 6.6 × 10-24 g) emitted by radium travels at 3.4 × 107± 0.1 × 107 mi/h. What is its de Broglie wavelength (in meters)?
wavelength = h/mv
To find the de Broglie wavelength of the alpha particle, we can use the de Broglie wavelength equation:
λ = h / p,
where
λ is the de Broglie wavelength,
h is Planck's constant (6.626 × 10^(-34) J·s),
p is the momentum of the particle.
The momentum of the alpha particle can be calculated using the formula:
p = m * v,
where
m is the mass of the particle (6.6 × 10^(-24) g),
v is the velocity of the particle (3.4 × 10^7 ± 0.1 × 10^7 mi/h).
First, we need to convert the mass to kilograms:
m = 6.6 × 10^(-24) g = 6.6 × 10^(-27) kg.
Next, we need to convert the velocity from miles per hour (mi/h) to meters per second (m/s). To do this, we'll use the conversion factor:
1 mi/h = 0.44704 m/s.
v = (3.4 × 10^7 ± 0.1 × 10^7) mi/h = (3.4 × 10^7 ± 0.1 × 10^7) * 0.44704 m/s.
Now, we can calculate the momentum:
p = m * v = (6.6 × 10^(-27) kg) * [(3.4 × 10^7 ± 0.1 × 10^7) * 0.44704 m/s].
After calculating the value of p, we can substitute it into the de Broglie wavelength equation to find the de Broglie wavelength λ.
Note: The ± symbol in the velocity indicates that the velocity is given with uncertainty. The uncertainty will be propagated through the calculations to determine the uncertainty in the de Broglie wavelength.