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JEE Advanced 2016 Paper 1 · Chemistry · Chemical Kinetics · easy

The radioactive isotope ^23592U finally changes to the stable isotope ^20782Pb through a series of α and β^- emissions. If x is the number of α…

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The radioactive isotope 92235U^{235}_{92}\mathrm{U} finally changes to the stable isotope 82207Pb^{207}_{82}\mathrm{Pb} through a series of α\alpha and β−\beta^- emissions. If xx is the number of α\alpha particles and yy is the number of β−\beta^- particles emitted per atom, find x−yx-y.
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Answer: 3

Only α\alpha emission changes the mass number, and each α\alpha lowers it by 44 (and the atomic number by 22). The mass number falls from 235235 to 207207, a drop of 2828, so x=284=7x=\dfrac{28}{4}=7. The 77 alpha particles lower the atomic number by 1414, to 92−14=7892-14=78. Each β−\beta^- emission raises the atomic number by 11 (a neutron changes to a proton) without changing the mass number, so reaching Z=82Z=82 needs y=82−78=4y=82-78=4 beta particles. Check on charge: 92−2(7)+4=8292-2(7)+4=82, as required for lead, and the mass check 235−4(7)=207235-4(7)=207 also holds. Therefore x−y=7−4=3x-y=7-4=3.
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