# Problem: The isotope 21884Po can decay by either α or β−emission.What is the energy release in each case? The mass of 21884Po, 21482Pb, 42He and 21885At is 218.008973 u, 213.999798 u, 4.002603 u and 218.008681 u, respectively.Express your answers using three significant figures separated by a comma.MeV for units

###### FREE Expert Solution

The energy released in the decay:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{m}}{{\mathbf{c}}}^{{\mathbf{2}}}}$, where m is the mass and c is the speed of light in a vacuum.

The reaction for α-decay can be expressed as:

$\overline{){}_{{\mathbf{84}}}{}^{{\mathbf{218}}}{\mathbf{P}}{\mathbf{o}}{\mathbf{\to }}{}_{{\mathbf{82}}}{}^{{\mathbf{214}}}{\mathbf{P}}{\mathbf{b}}{\mathbf{+}}{}_{{\mathbf{2}}}{}^{{\mathbf{4}}}{\mathbf{H}}{\mathbf{e}}}$

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###### Problem Details

The isotope 21884Po can decay by either α or β−emission.

What is the energy release in each case? The mass of 21884Po, 21482Pb, 42He and 21885At is 218.008973 u, 213.999798 u, 4.002603 u and 218.008681 u, respectively.