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Initially the nucleus of radium 226 is at rest. It decays due to which and `alpha` particle and the nucleus of radon are created. The released energy during the decay is `4.87` Mev, which appears as the kinetic energy of the two resulted particles `[m_(alpha)=4.002"amu",m_("Rn")=22.017"amu"]` Kinetic energies of `alpha` particle & radon nucleus are respectivelyA. `0.09 "Mev", 4.08 "Mev"`B. `4.78 "Mev", 0.09 "Mev"`C. `4.08 "Mev", 0.09 "Mev"`D. `3.68 "Mev", 0.08 "Mev"` |
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Answer» Correct Answer - B `P=mv` & `(P^(2))/(2m)=1/2mV^(2)` `E_(alpha)=(P^(2))/(2m)` `E_("radon")=(P^(2))/(2M)` Also `(P^(2))/(2m)+(P^(2))/(2M)=/_EimpliesP=sqrt(2/_E(mM)/(m+M))` `m=4.002"mu", M=222.017 "mu"` `m=6.64xx10^(-27)kg` |
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