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 6.40 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. m/s 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° clockwise from the +x-axis.) Direction ? Magnitude ?

          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 6.40  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.
 m/s
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° clockwise from the
+x-axis.)
Direction ?
Magnitude ?

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Added by Jacob G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Shyam P

04/03/2022

Step 1

Step 1: We know that the collision is elastic, which means both momentum and kinetic energy are conserved. Show more…

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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 6.40 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. m/s 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° clockwise from the +x-axis.) Direction ? Magnitude ?

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Transcript

00:01 Hello everyone in this problem we have given the connecting energy before the collision is 6 .40 into 10 raised to power minus 13 jowl.

00:13 Here i'm considering for gold it is 2 and for helium it is 1 and before collision i'm just writing that is b and for after collision i'm just writing small a.

00:27 So, mass of gold, that is m of 2 is 3 .29 into 10 days to power minus 25 kg.

00:40 Also, m1 is 6 .68 into 10 days to power minus 27 kg.

00:49 And we have given theta that is 120 degree.

00:54 Also the v of 2 before collision is 0 and we have to calculate the velocity of second particles after collision.

01:07 Now by conservation of energy low, the kinetic energy before collision should be equals to kinetic energy after collision.

01:20 So before collision, it is k of eb is equals to after collision half of m1 v1a square plus half of m2 v2a square.

01:37 Now as also we know that 1 of b minus 1 of 2b is equals to v of 2a minus v of 1a.

01:51 So value of v2a comes out v of 1b plus v of 1a.

01:59 So use this value in this equation we get k of e v is equal to half of m1 v1a square plus half of m2 into v1b plus v1a whole square.

02:22 So after solving the whole square formula we get m1 plus m2 into v1a square plus 2 plus 2 v1a 2 times m2 v1b into v1a plus m2 v1b square minus 2 k eb is equals to 0.

02:50 Now we know these form values...

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