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 7.87 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 141° 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 Incorrect: Your answer is incorrect. m/s 197Au velocity Incorrect: Your answer is incorrect. m/s 197Au direction Incorrect: Your answer is incorrect. ° counterclockwise from the +x-axis (b) What is the final kinetic energy (in J) of the helium nucleus? Incorrect: Your answer is incorrect. J

Explanation / Answer

Vi ( inital velocity of Hekium) =sqroot ( 2E/m) = sqroot ( 2x7.87 x 10^-13 / 6.68 x10^-27) = sqroot (2.356 x 10^14) = 1.535 x 10^ 7 m/s

conserving the momentum along x axis,

Vf( for helium) x =(6.68 1027 - 3.29 1025 ) 1.535 x 10^7/ ( 6.68 1027 + 3.29 1025) = - 4.947x 10^ 7 /3.3568 = -1.473 x 10^7 m/s apprx

Final speed = 1.473 x 10^ 7 m/s for He

Vy ( y component of helium's velocity) = 1.473 sin 141= 0.927 x 10^ 7 m/s

vx = 1.473 cos 141= -1.1447 m/s apprx

Pxi = Pxf ( He) + Pxf ( Au)

6.68 1027 ( 1.535 x 10^ 7 ) = 6.68 x 10^-27 (1.1447 x 10^ 7 ) + 3.29 1025 ( Vx)

Vx ( for Au) = 0.792 x 10^ 5 m/s

Pyf ( He) = Pyf ( Au)

0.927 x 10^ 7 m/s (6.68 1027) = 3.29 1025 ( Vf for Au)

Vf = 1.882 x 10^ 5 m.s

V ( for Au) = 2.04 x10^5 m/s

angle = tan^-1 ( 1.882/ 0.792) =67.18 degree apprx

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