Question

Which of the following is/are correct when a nuclide of mass number (A) and atomic number (Z) undergoes radioactive process
This question has multiple correct options
A.Both A and Z decrease, the process is called
B.A remains Beta plus(or) positron decay (or) K-electron
C.Both A and Z remain unchanged, the process is called as gamma-decay
D.Both A and Z increase, the process is called nuclear isomerism

Answer 1

We know that radioactivity takes place when the nucleus of an unstable atom loses energy by emitting energy in form of electromagnetic waves (or) emitted particles known as radiation. In simple words, the immediate emission of radiation in the form of high energy photons (or) particles rising from nuclear reaction is known as radioactivity. Radioactive decay, nuclear disintegration, nuclear decay are the other words of radioactivity.

Complete step by step answer:
We must remember that the decay of a nucleus by emitting an alpha particle $\left( {\alpha \,\left( {or} \right)\,{}_2^4He} \right)$ is called alpha emission. The general balanced equation for the alpha emission is given by,
${}_Z^AX \to {}_2^4\alpha + {}_{\left( {Z - 2} \right)}^{\left( {A - 4} \right)}Y$
Where,
$A$ is the mass number.
$Z$ is the atomic number.
$X$ is the original isotope.
$Y$ is the new isotope.
We can see that both A and Z decrease. Therefore, the option (A) is correct.
We have to know that the decay of a nucleus by emitting a positron $\left( {{{\text{e}}^{\text{ + }}}} \right)$ is called as positron emission. In positron emission, one proton of the original nucleus decays to a form positron and a neutron. The new nucleus formed has one fewer proton and one more neutron than the original nucleus.
When a nucleus emits a positron, the mass number remains unchanged and the atomic number reduces by one.
The general equation is,
${}_Z^AX \to {}_{ + 1}^0e + {}_{\left( {Z - 1} \right)}^AY$
Where,
$A$ is the mass number.
$Z$ is the atomic number.
$X$ is the original isotope.
$Y$ is the new isotope.
We can see that the mass number remains constant, but there is a change in the atomic number.
Therefore, the option (B) is correct.
We must remember that the gamma rays are liberated out by many radioactive sources besides alpha or beta particles. After emission of alpha or beta particles the remaining nucleus may still be in an excited energy state by giving out a gamma photon it decreases to a lower energy state. Gamma rays contain no electrical charge related with them. In gamma decay, there is no change in the atomic number and mass number.
Therefore, the option (C) is correct.
As we know nuclear isomerism does not take place when there is an increase in the values of atomic number and mass number.
Therefore, the option (D) is incorrect.

So, the correct answer is Option A,B,C.

Note: We must remember that the decay of nucleus by emitting a beta $\left( {{{\text{e}}^{\text{ - }}}} \right)$ particle is called as beta emission. In beta emission, one neutron of the original nucleus decays to form a beta particle and a proton. Due to which, the new nucleus formed has one more proton and one fewer neutron than the original nucleus.
The general equation is given as,
${}_Z^AX \to {}_{ - 1}^0e + {}_{\left( {Z + 1} \right)}^AY$
Where,
$A$ is the mass number.
$Z$ is the atomic number.
$X$ is the original isotope.
$Y$ is the new isotope.