Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in chem

Question

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 (197Au). See the figure below. The energy of the incoming helium nucleus was 9.70 10-13 J, and the masses of the helium and gold nuclei were 6.68 10-27 and 3.29 10-25 kg, respectively (note that their mass ratio is 4 to 197). If a helium nucleus scatters to an angle of 120° during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed.

Explanation / Answer

KE = m v0^2 / 2

9.70 x 10^-13 = (6.68 x 10^-27) v0^2 /2

v0 = 1.70 x 10^7 m/s

Applying momentum conseration,

(6.68 x 10^-27)(1.70 x 10^7)i + 0 = (6.68 x 10^-27)(v1)(cos120i + sin120j) + (3.29 x 10^-25)v2(cos(theta)i + sin(theta)j)

1.1356 x 10^-19i = (-3.34v1 x 10^-27 + 3.29 x 10^-25 v2 cos(theta)i + (5.785 x 10^-27v1 + 3.29 x 10^-25 v2 sin(theta))j

v2 cos(theta) = 345000 + 0.01 v1 ... (i)

v2 sin(theta) = - 0.0176v1 .... (ii)

(i)^2 + (ii)^2 => v2^2 =(4.09 x 10^-4)v1^2 + 690v1 + 1.19 x 10^11

Applying energgy conseration,

9.70 x 10^-13 = 3.34 x 10^-27 v1^2 + 1.645 x 10^-25 v^2

v1 = 1.67 x 10^7 m/s and v2 = 471 m/s

Ans: 1.67 x 10^7 m/s

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