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 nuclei (197Au). The energy of the incoming helium nucleus was 7.97 1013 J, and the masses of the helium and gold nuclei were 6.68 1027 kg and 3.29 1025 kg, respectively (note that their mass ratio is 4 to 197). (Assume that the helium nucleus travels in the +x direction before the collision.)

(a)

If a helium nucleus scatters to an angle of 121° during an elastic collision with a gold nucleus, calculate the helium nucleus' final speed (in m/s) and the final velocity (magnitude in m/s and direction counterclockwise from the +x-axis) of the gold nucleus.

4He speed ______ m/s

197Au velocity ______ m/s

197Au direction ______ ° counterclockwise from the +x-axis

(b)

What is the final kinetic energy (in J) of the helium nucleus?

_____ J

Explanation / Answer

a)
1/2m1v1^2 = KEi
=> vi = sqrt(2KEi/mi)
=> vi = sqrt((2*7.97*10^-13)/(6.68*10^-27))
=> vi = 1.54*10^7 m/s

we have
vi = 1.54*10^7 m/s
theta = 121 deg
m1 = 6.68*10^-27 kg
m2 = 3.29*10^-25 kg

v1' = (-3.237*10^5 + sqrt((3.237*10^5)^2 - 4(1.0203)(-2.348*10^14)))/(2*1.0203)
= 1.5*10^7 m/s (4He speed )
v2' = 5.36*10^5 m/s (197Au velocity)

theta = arctan(-0.56529)
= -29.5 deg (197Au direction)

b)
KEf = 0.5m1v1'^2

= 0.5*6.68*10^-27*(1.5*10^7)^2
= 7.5*10^-13 J

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