Question

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass m

Answer 1

Answer:

$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$

Explanation:

$\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 }$$\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }$$\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}$$\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }$

$\text{So, from the above infromation;}$

$radius (r) = \dfrac{15 \ fm }{2}$

$radius (r) = 7.5 \times 10^{-15} \ m$

$\text{Using the formula for potential energy}$

$U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}$

$U = 5.529 \times 10^{-12} \ J$

$\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}$$\dfrac{1}{2}mv_a^2= 5.529 \times 10^{-12} J$

$v_a = \sqrt{\dfrac{2(5.529 \times 10^{-12} \ J)}{4(1.67 \times 106{-37} \ kg)}}$

$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$