Question

If a helium nucleus scatters at an angle of 120° from the initial direction of motion during an elastic collision with a gold nucleus, calculate the helium nucleus’s final speed, in meters per second. Ernest Rutherford (who was the first person born in New Zealand 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). During this experiment, the energy of the incoming helium nucleus was 8.00 × 10-13 J, and the masses of the helium and gold nuclei were 6.68 × 10-27 kg and 3.29 × 10-25 kg, respectively (note that their mass ratio is 4 to 197).

Answer 1

Answer:

the helium nucleus’s final speed = $1.55*10^7 \ m/s$

Explanation:

From the question; we are to calculate just only the helium nucleus’s final speed, in meters per second.

From the  knowledge of Kinetic energy;

Kinetic energy K.E = $\frac{1}{2} m_1v_1^2$

Making the velocity the subject  of the formula from the above expression; we have:

$v_1 = (\frac{2K.E}{m_1})^{\frac{1}{2}$

where;

K.E = 8.00 × 10⁻¹³ J

m₁ =  6.68 × 10⁻²⁷ kg

$v_1 = (\frac{2*8.00*10^{-13}}{6.68*10^{-27}} )^{({\frac{1}{2})}$

$v_1 = 1.55*10^7\ m/s$