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

To find the kinetic energy of the particle, we need to use the de Broglie wavelength equation and the equation for kinetic energy.

The de Broglie wavelength (λ) is given by:

λ = h / p

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

In this case, the de Broglie wavelength is given as equal to the diameter of the target nucleus, which is 1.4 x 10^-14 m.

Therefore, we can equate the de Broglie wavelength to the diameter:

1.4 x 10^-14 m = h / p

Next, we can rearrange the equation to solve for momentum:

p = h / (1.4 x 10^-14 m)

Now, we need to find the magnitude of momentum (p), which is equal to the product of mass (m) and velocity (v):

p = m * v

Given that mass (m) is 6.64 x 10^-27 kg, we can substitute this into the equation:

6.64 x 10^-27 kg * v = h / (1.4 x 10^-14 m)

Solving for velocity (v):

v = (h / (1.4 x 10^-14 m)) / (6.64 x 10^-27 kg)

Now, we can calculate the velocity of the particle.

Finally, we can find the kinetic energy (KE) using the equation:

KE = 0.5 * m * v^2

Substituting the mass (m) and the velocity (v) calculated above, we can find the value of the kinetic energy.

Note: The calculation assumes non-relativistic speeds and ignores relativistic effects as mentioned in the question.