The Energy of an Alpha Particle with a De Broglie Wavelength of 0.004 Angstroms

To determine the energy of an alpha particle with a de Broglie wavelength of 0.004 angstroms, we can use the de Broglie equation, which relates the wavelength (λ) to the momentum (p) of a particle:

λ = h / p

where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J*s), and p is the momentum of the particle.

From the equation, we can solve for the momentum of the alpha particle:

p = h / λ

Step 1: Convert the de Broglie wavelength to meters
To use the equation, we need to convert the de Broglie wavelength from angstroms to meters. 1 angstrom is equal to 10^-10 meters, so the wavelength can be expressed as:

λ = 0.004 x 10^-10 meters

Step 2: Calculate the momentum
Now that we have the wavelength in meters, we can calculate the momentum of the alpha particle:

p = 6.626 x 10^-34 J*s / (0.004 x 10^-10 meters)

Simplifying this expression gives us:

p = 1.6565 x 10^-25 kg*m/s

Step 3: Calculate the energy
The energy of a particle can be determined using the equation:

E = p^2 / (2m)

where E is the energy, p is the momentum, and m is the mass of the particle.

The mass of an alpha particle is approximately 6.644 x 10^-27 kg.

Calculating the energy using the given values:

E = (1.6565 x 10^-25 kg*m/s)^2 / (2 * 6.644 x 10^-27 kg)

Simplifying the expression results in:

E = 3.268 x 10^-12 J

Therefore, the energy of an alpha particle with a de Broglie wavelength of 0.004 angstroms is 3.268 x 10^-12 Joules.

In conclusion, by applying the de Broglie equation and using the given de Broglie wavelength of 0.004 angstroms, we were able to calculate the energy of the alpha particle to be 3.268 x 10^-12 Joules.