Question

The radioactive element radium (Ra) decays by a process known asalpha decay,in which the nucleus emits a helium nucleus. (Factoid: These high-speed helium nuclei were named alpha particles when radioactivity was first discovered, long before the identity of the particles was established.) The nuclear reactionthat occursis226Ra→222Rn+4He,where Rn is the element radon. The accurately measured atomic masses of the three atoms are 226.0254 u, 222.0176 u, and 4.0026 u. How much energy is released in each decay? (The energy released in radioactive decay is what makes nuclear waste "hot.")

Answer 1

Answer: The energy released in the decay process is
`4.6800* 10^(11)J`

Step-by-step explanation:

The equation for the alpha decay of Ra-226 follows:

`_(88)^(226)\textrm{Ra}\rightarrow _(2)^(4)\textrm{He}+_(86)^(222)\textrm{Rn}`

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

`\Delta m=(m_(Rn)+m_(He))-(m_(Ra))`

We know that:

`m_(Rn)=222.0176u\\m_(Ra)=226.0254u\\m_(He)=4.0026u`

Putting values in above equation, we get:

`\Delta m=(222.0176+4.0026)-(226.0254)=-0.0052g=-5.2* 10^(-6)kg`

(Conversion factor: 1 kg = 1000 g )

To calculate the energy released, we use Einstein equation, which is:

`E=\Delta mc^2`

`E=(-5.2* 10^(-6)kg)* (3* 10^8m/s)^2`

`E=-4.6800* 10^(11)J`

Hence, the energy released in the decay process is
`4.6800* 10^(11)J`