# 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.