Question and Answers Forum

All Questions      Topic List

Moderm Physics Questions

Previous in All Question      Next in All Question      

Previous in Moderm Physics      Next in Moderm Physics      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com