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Question

A radioactive element 90X238 decays in to 82Y222. The number of β-particles emitted is:

A
0
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B
4
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C
6
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D
1
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Solution

The correct option is A 0
Given- Radioactive element 90X23882Y222+ particles

we know that - * For 1 α particle emission mass reduces by (4) Atomic num. reduces by (2) . * For 1 β particle emission No change in mass Atomic number increases by (1).

For 90X23882Y222
Change in mass num =16

we can say that (4) α-particles are released.
since, β - barticles bring no change in mass.
Hence, 4 alpha particles reduces the atomic number
by a value of (8).


And in the decay exactly atomic number reduces by ( 8 ) . Hence, 4α -particles are released and 0β -particles are emitted


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