Question

41. An element P decays into element R by a multi step
process P Q + 2Het and Q R + 2e-. Then
(1) P and Rare isotopes A
27 P and Rare isobarst
(3) Q and R are isotopes
(4) P and Q are isobars
2​

Answer 1

Answer:

Since P and R is of same atomic number so correct answer is

(1) P and Rare isotopes

Explanation:

As we know by the given equations

$P --> Q + _2^4He$

$Q --> R + 2e^-$

now we know from above equation that first equation is alpha decay while the other equation is beta decay

so in first equation

element P is having

atomic number z and mass number A

element Q

atomic number z - 2 and mass number A - 4

in second equation it is beta decay

element R

atomic number z and mass number A - 4

So here we can say that P and R is of same atomic number so they are isotopes while Q and R is of same mass number so they are isobars

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Topic : isotopes

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