In one of the classic nuclear physics experiments at the beginning of the 20th century, alpha particles were accelerated toward stationary gold nuclei. Some alpha particles were reflected straight back due to the Coulomb interaction between the alpha particle and the nucleus. If the kinetic energy of a doubly charged alpha nucleus was 3.83 MeV when it was a large distance away from the nucleus, how close to the gold nucleus (79 protons) could it come before being reflected

Explanation:

In the whole process , electric potential energy is converted into kinetic energy .

Kinetic energy = 3.83 MeV

= 3.83 x 1.6 x 10⁻¹⁶ J

= 6.128 x 10⁻¹⁶ J .

Let the closest distance of approach be r .

Electric potential energy = k Q q / r , Q is charge on nucleus , q is charge on alpha particle , r is closest distance .

Electric potential energy = 9 x 10⁹ x 79 x 1.6 x 10⁻¹⁹ x 2 x 1.6 x 10⁻¹⁹ / r

= 3640.32 x 10⁻²⁹ / r

So,

6.128 x 10⁻¹⁶= 3640.32 x 10⁻²⁹ / r

r = 3640.32 x 10⁻²⁹ / 6.128 x 10⁻¹⁶

= 594.05 x 10⁻¹³

= 59.405 x 10⁻¹²

= 59.405 pm .

Answer

a)= negative

b)= c

c)= B, there is no correlation

Step-by-step explanation: