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 6.10 Times 10^-13 J, and the masses of the helium and gold nuclei were 6.68 Times 10^-27 and 3.29 Times 10^-25 kg, respectively (note that their mass ratio is 4 to 197). If a helium nucleus scatters to an angle of 120 degree during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed. Calculate the final velocity (magnitude and direction) of the gold nucleus. (Assume the positive x direction is the direction in which the helium nucleus is initially traveling, and that it scatters 120 degree clockwise from the +x-axis.)

Explanation / Answer

m1 V1=m1 V '1 + m2 V2

Speeds introduced as vectors:

Velocity of helium= sqrt( 2U/ m) = sqrt (2*6.10X10-13 /6.68X10-27 ) = 13.51 X106 m/s

6.68X10-27( 13.51X106 m/s) i = 6.68X10-27(12900000) ( -cos(60) i - sin(60) j )+ 3.29X10-25 V2

This gives us that:

V2= 405767.1733 i + 226830.2161 j

The magnitude of the velocity is the square root of the sum of th x and y components:

V2= 464864.4382 m/s

The angle of the direction is:

tan(theta) = 226830.2161/405767.1733

theta= 29.21 degrees measured counter clockwise

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