Question

A radioactive element ₉₀X²³⁸ decays into ₈₃Y ²²² . The number of b -particles emitted are 

(a) 1 

(b) 2 

(c) 4 

(d) 6

Answer 1

Answer is (a) 1

α-particles are emitted = \(\frac{238 - 222}{4}\) = 4

The atomic number is decreased 90 - 4 x 2 = 82

As atomic number of ₈₃Y²²².

So, atomic number is increased by 1, therefore one β -particle is emitted.