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A-nuclide-81-210-X-decays-to-another-nuclide-80-A-Y-in-four-successive-radioactive-decays-Each-decay-involves-the-emmision-of-either-an-alpha-particle-or-a-beta-particle-The-value-of-A-is-




Question Number 137429 by physicstutes last updated on 02/Apr/21
A nuclide _(81)^(210) X decays to another nuclide _(80)^A Y in   four successive radioactive decays. Each decay  involves the emmision of either an alpha particle  or a beta particle. The value of A is:  A. 120           B. 206  C. 208            D. 212
Anuclide81210Xdecaystoanothernuclide80AYinfoursuccessiveradioactivedecays.Eachdecayinvolvestheemmisionofeitheranalphaparticleorabetaparticle.ThevalueofAis:A.120B.206C.208D.212
Commented by Dwaipayan Shikari last updated on 02/Apr/21
X^(210) _(81) →^(−α) H_(79) ^(206)   H_(79) ^(206) →^(−β_0 ) Y_(80) ^(206)   A=206
X81210αH79206H79206β0Y80206A=206
Commented by physicstutes last updated on 02/Apr/21
sir does this two decay not sum up  to two. its four they asked for.
sirdoesthistwodecaynotsumuptotwo.itsfourtheyaskedfor.
Commented by Dwaipayan Shikari last updated on 02/Apr/21
Is there any γ decay sir?
Isthereanyγdecaysir?
Commented by physicstutes last updated on 02/Apr/21
no just alpha and beta
nojustalphaandbeta

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