Question

an alpha particle (mass=6.6x10^-24g) emmited by a radium isotope travels at 3.4x10^7 mi/hr. what is its de broglie wavelength in meters

Answer 1

The de Broglie wavelength of the alpha particle is approximately 6.63 x 10^-16 m.

Step-by-step explanation:

To calculate the de Broglie wavelength of an alpha particle, we first need to convert its speed from miles per hour to meters per second. Given that the speed of light is approximately 3 x 10^8 m/s, the conversion factor is 0.447 x 10^8 m/s. Therefore, the speed of the alpha particle is approximately 1.5188 x 10^8 m/s.

Then, we can use the de Broglie wavelength formula:

λ = h / p

Where λ is the de Broglie wavelength, h is Planck's constant (approximately 6.626 x 10^-34 J·s), and p is the momentum of the particle.

The momentum of the alpha particle can be calculated using the formula:

p = m * v

Where p is the momentum, m is the mass of the particle, and v is its velocity.

Substituting the given values:

p = (6.6 x 10^-24 g) * (1.5188 x 10^8 m/s)

Converting the mass to kilograms (1 g = 10^-3 kg), we get:

p = (6.6 x 10^-27 kg) * (1.5188 x 10^8 m/s)

Calculating the product gives us:

p = 9.99 x 10^-19 kg·m/s

Now we can calculate the de Broglie wavelength:

λ = (6.626 x 10^-34 J·s) / (9.99 x 10^-19 kg·m/s)

Simplifying the expression:

λ = 6.63 x 10^-16 m

Therefore, the de Broglie wavelength of the alpha particle is approximately 6.63 x 10^-16 m.

Answer 2

We know that E = 0.5mv^2 = hc/(lambda), where lambda is the de broglie wavelength. E = 0.5mv^2 due to the speed that the alpha particle possess in the form of kinetic energy. Hope this helps!