Answer:
the helium nucleus’s final speed = [tex]1.55*10^7 \ m/s[/tex]
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 = [tex]\frac{1}{2} m_1v_1^2[/tex]
Making the velocity the subject of the formula from the above expression; we have:
[tex]v_1 = (\frac{2K.E}{m_1})^{\frac{1}{2}[/tex]
where;
K.E = 8.00 × 10⁻¹³ J
m₁ = 6.68 × 10⁻²⁷ kg
[tex]v_1 = (\frac{2*8.00*10^{-13}}{6.68*10^{-27}} )^{({\frac{1}{2})}[/tex]
[tex]v_1 = 1.55*10^7\ m/s[/tex]
