# In a Rutherford scattering experiment a target nucleus has a diameter of 1.4 10-14 m. The incoming has a mass of 6.64 10-27 kg. What is the kinetic energy of an particle that has a de Broglie wavelength equal to the diameter of the target nucleus? Ignore relativistic effects

The momentum is

p = h/lambda

Kinetic energy is:

p^2/(2m) = (h/lambda)^2/(2m) =

8.43*10^(-14)J = 0.526 MeV.

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## To find the kinetic energy of the particle that has a de Broglie wavelength equal to the diameter of the target nucleus, you can follow these steps:

1. Use the de Broglie wavelength equation: λ = h/p, where λ is the wavelength, h is the Planck's constant (6.63 × 10^-34 J·s), and p is the momentum of the particle.

2. Rearrange the equation to solve for the momentum: p = h/λ.

3. Substitute the given diameter of the target nucleus (1.4 × 10^-14 m) as the de Broglie wavelength in the equation: λ = 1.4 × 10^-14 m.

4. Calculate the momentum of the particle using the equation from step 2.

5. Use the kinetic energy equation: KE = p^2 / (2m), where KE is the kinetic energy and m is the mass of the particle.

6. Substitute the momentum calculated in step 4 and the given mass of the incoming particle (6.64 × 10^-27 kg) into the kinetic energy equation.

7. Calculate the kinetic energy using the equation from step 5.

By following these steps, you will find that the kinetic energy of the particle is approximately 8.43 × 10^-14 J or 0.526 MeV (mega-electron volts).