KVPY SA Exam 2011 Question Paper

Question : 29 of 80

Marks: +1, -0

A nucleus of lead

Pb^214_82

emits two electrons followed by an alpha particle. The resulting nucleus w ill have

Solution:

\Pb {_82↖{214}} {–––––––– ▶}↙{2e^{-}+ {_2\He^4}}X{_82↖{210}}

82 → Proton
210 – 82 = 128 Neutron

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