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In the final Uranium radioactive series the initial nucleus is `U_(92)^(238) ` and the final nucleus is `Pb_(82)^(206)` . When Uranium neucleus decays to lead , the number of a - particle is …….. And the number of `beta` - particles emited is …… |
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Answer» Correct Answer - `8alpha and 6 beta` Let x be the number of alpha particles and y be the no. of `beta` particles emitted in the disintegration process of `._(92)U^(238)` to `._(82)Pb^(206)`. We may write `._(92)U^(238) to._(82)Pb^(206)+(._(2)He^(4))+y(._(-1)e^(0))` form law of conservation of mass number `238=206+4x+y(0)=206+4x` `=4x=238-206=32, x=8` form law of conservation of mass number, `92=82+2x-y=-10+2x` `y=-10+2(8)=6` Hence `8alpha` particles and `6 beta` particles are emitted. |
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