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

`\mathbf{v_a = 4.06 * 10^7 \ m/s}`

Step-by-step 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) = (15 \ fm )/(2)`

`radius (r) = 7.5 * 10^(-15) \ m`

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

`U = ((9* 10^9 \ Nm^2/C^2)(2(1.6*10^(-19) \ C ))(90)(1.6* 10^(-19) C))/(7.5 * 10^(-13) \ m)`

`U = 5.529 * 10^(-12) \ J`

`\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}`
`(1)/(2)mv_a^2= 5.529 * 10^(-12) J`

`v_a = \sqrt{\frac{2(5.529 * 10^(-12) \ J)}{4(1.67 * 106{-37} \ kg)}}`

`\mathbf{v_a = 4.06 * 10^7 \ m/s}`