Enter your answer in scientific notation. In 1911, Ernest Rutherford discovered

Question

Enter your answer in scientific notation. In 1911, Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. Ira helium nucleus with a mass of5.68 x 10-27 kg, a charge of +2e, and an initial velocity of 1.49 × 107 m/s is projected head-on toward a gold nucleus with a charge of +79e, how close will the helium atom come to the gold nucleus before it stops and turns around? Assume the gold nucleus is held in place by other gold atoms and does not move. x 10 (select) m.

Explanation / Answer

mass of helium, m = 6.68*10^-27 kg
charge, q = 2e = 2*1.6*10^-19 C
initial velocity vo = 1.49*10^7 m/s

charge on gold nucleus, Q = 79e

let distance of closest approach be d

then initial KE of helium atom = 0.5m(vo)^2
Final PE at distance of closest approach = kQq/d [ k is coloumbs constant]
from conservation of energy
8.98*10^9*79*2*(1.6*10^-19)^2/d = 0.5*6.68*10^-27*(1.49*10^7)^2
d = 4.894*10^-14 m

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