Question
Find the energy (in MeV) released when $\alpha$ decay converts radium $\frac{226}{88} \mathrm{Ra}$ (atomic mass $=226.02540$ u) into radon $\frac{222}{86} \mathrm{Rn}$ (atomicmass $=222.01757 \mathrm{u} )$ The atomic mass of an $\alpha$ particle is 4.002603 $\mathrm{u}$
Step 1
The initial mass is the atomic mass of radium, which is $226.02540$ u. The final mass is the sum of the atomic mass of radon and the atomic mass of an $\alpha$ particle, which is $222.01757$ u + $4.002603$ u = $226.020173$ u. Show more…
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$$ \text { Find the energy (in MeV) released when } \alpha \text { decay converts radium } $$ $\frac{226}{88} \mathrm{Ra}$ (atomic mass $\left.=226.02540 \mathrm{u}\right)$ into radon ${ }_{86}^{222} \mathrm{Rn}$ (atomic mass $=$ $222.01757 \mathrm{u}) .$ The atomic mass of an $\alpha$ particle is $4.002603 \mathrm{u}$.
Find the energy (in MeV) released when $\alpha$ decay converts radium 226 Ra (atomic mass $=226.02540 \mathrm{u}$ ) into radon $\frac{222}{86} \mathrm{Rn}($ atomic mass $=222.01757 \mathrm{u}) .$ The atomic mass of an $\alpha$ particle is $4.002603 \mathrm{u}$.
Find the energy (in MeV) released when $\alpha$ decay converts radium ${ }_{88}^{226} \mathrm{Ra}$ (atomic mass $=226.02540 \mathrm{u}$ ) into radon ${ }_{86}^{222} \mathrm{Rn}$ (atomic mass $=222.01757 \mathrm{u}$ ). The atomic mass of an $\alpha$ particle is $4.002603 \mathrm{u}$.
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