In Rutherford's famous experiment,alpha particles of mass 6.64 * 10^-27 kg and charge + 2e were fired directly at a gold nucleus (charge + 79e)with a speed of 3 * 10⁷ m/s.Due to the Coulomb repulsive force the alpha particles slowed down and eventually came to rest momentarily before turning around.The distance alpha particles came closest to the gold nucleus before turning around was

Question

Quizplus · Accepted Answer

In Rutherford's experiment, the closest approach distance (r) can be calculated using the equation:

r = (kQq)/mv^2

where k is Coulomb's constant, Q and q are the charges of the gold nucleus and the alpha particle, m is the mass of the alpha particle, and v is its speed.

Substituting the given values, we get:

r = [(9x10^9 Nm^2/C^2) x (2e) x (79e)] / [(6.64x10^-27 kg) x (3x10^7 m/s)^2]

r = 12.9 x 10^-15 m = 12.9 fm

Therefore, the closest approach distance of the alpha particles to the gold nucleus before turning around was 12.1 fm (approximately). Thus, the best choice is C.

In Rutherford's Famous Experiment,alpha Particles of Mass 6