Question

A radioactive element undergoes 2alpha decays and 3beta decays successively. What will be the change in atomic and mass number​

Answer 1

When a radioactive nucleus emits 1 α-particle mass number decreases by 4 and the atomic number decreases by 2.

So after the emission of 3−α particles

New atomic mass, A

′

=A−4×3=A−12

New atomic number Z

′

=Z−6

When this nucleus emits 1β-particle (positron), the atomic mass remains unchanged but the atomic number decreases by 1.

So after emission of 2 positrons

A"=A

=A−12

Z"=Z

−2×1=Z−8

So, the ratio of neutrons to protons-

p

n=

Z−8

A−Z−4

Answer 2

Explanation:

In alpha decay, the atomic number of parent nucleus decreases by 2.

In beta decay, the atomic number of parent nucleus increases by 1.

Whereas the atomic number of the parent nucleus remains the same in gama decay.

Reaction :

Z

A

X →

Z−2

A−4

Y →

Z−1

A−4

W

∗

→

Z−1

A−4

W

Thus overall the atomic number of daughter is less than parent atom by 1.