The radioactive element radium (Ra) decays by a process known as alpha decay, in

Question

The radioactive element radium (Ra) decays by a process known as alpha decay, in which the nucleus emits a helium nucleus. (These high-speed helium nuclei were named alpha particles when radioactivity was first discovered, long before the identity of the particles was established.) The reaction is 226Ra222Rn+4He, where Rn is the element radon. The accurately measured atomic masses of the three atoms are 226.025 u , 222.017u , and 4.003u , where u=1.66×1027 kg.

A. How much energy is released in each decay? (The energy released in radioactive decay is what makes nuclear waste "hot.")

E=? Units=?

Explanation / Answer

Solution-   226Ra222Rn+4He + mass lost

So, mass lost = 226.025 u - (222.017u + 4.003u )

mass lost = 0.005 u = 0.005 x 1.66×1027 kg = 8.3 x 10^-30 kg

Now, E = mc^2 = 8.3x10^-30 kg x (3 x10^8 m/s)^2 =7.47 x 10^-13 J

That's energy per decay of one atom. You can multiply by avogadro's number for decay per mol of atoms.

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