Answer:
[tex]\mathbf{v_a = 4.06 \times 10^7 \ m/s}[/tex]
Explanation:
[tex]\text{The missing part of the question is attached below.} \\ \\ \text{ mass m = 4u and charge q = 2e. Suppose a uranium nucleus with 92 protons decays into }[/tex][tex]\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }[/tex][tex]\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}[/tex][tex]\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }[/tex]
[tex]\text{So, from the above infromation;}[/tex]
[tex]radius (r) = \dfrac{15 \ fm }{2}[/tex]
[tex]radius (r) = 7.5 \times 10^{-15} \ m[/tex]
[tex]\text{Using the formula for potential energy}[/tex]
[tex]U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}[/tex]
[tex]U = 5.529 \times 10^{-12} \ J[/tex]
[tex]\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}[/tex][tex]\dfrac{1}{2}mv_a^2= 5.529 \times 10^{-12} J[/tex]
[tex]v_a = \sqrt{\dfrac{2(5.529 \times 10^{-12} \ J)}{4(1.67 \times 106{-37} \ kg)}}[/tex]
[tex]\mathbf{v_a = 4.06 \times 10^7 \ m/s}[/tex]
