92U238 decays successively to form 90Th234,91Pa234,92U234 . The sequence of radiations emitted is
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a
β,α,β
b
α,β,α
c
α,β,β
d
β,β,α
answer is C.
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Detailed Solution
92U238→α90Th234→β91Pa234→β92U234when a nucleus emits an alpha particle its mass number decreases by 4 and atomic number decreases by 2.when a nucleus emits a beta particle its atomic number increases by 1.