Question

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.64×10-27 kg. What is the kinetic energy of an α particle that has a de Broglie wavelength equal to the diameter of the target nucleus? Ignore relativistic effects.

Answer 1

E = 2.5 x 10⁻¹⁴ J

Step-by-step explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

`\lambda = (h)/(mv)`

`1.33 * 10^(-14)= (6.626 * 10^(-34))/(6.64 * 10^(-27)* v)`

`v= (6.626 * 10^(-34))/(6.64 * 10^(-27)* 1.33 * 10^(-14))`

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

`E = (1)/(2)mv^2`

`E = (1)/(2)* 6.64 * 10^(-27)* (7.5* 10^6)^2`

E = 2.5 x 10⁻¹⁴ J