91Ó°ÊÓ

Any species of atom that persists for a quantifiable period is referred to as a nuclide.

a)

A: a large number

Z is the atomic number (number of protons)

N: the total number of neutrons (A - Z)

Beta decay: nuclides with a high N/Z ratio (above the stability zone) will undergo beta decay. When a neutron decays into a proton, N value decreases, Z value increases, and the N/Z ratio decreases.

Positron emission/electron capture: nuclides with a low N/Z ratio (below the stability zone) will emit positrons or collect electrons. As a result of the proton being converted to a neutron N increases Z decreases, and the N/Z ratio increases.

Alpha decay: nuclides with Z>83 (too heavy) will undergo alpha decay, so both N and Z values will decrease by 2 (and become lighter).

${}_{92}{}^{238}\mathbf{U}$

Here one can see that the atomic number of U is 92(>83) which makes it too heavy to be stable, so, these nuclei will undergo alpha decay.

b)

${}_{24}{}^{48}\text{Cr}$

The number of protons is Z = 24.

The number of neutrons is;

$\begin{array}{rcl}N& =& A-\text{Z}\\ & =& 48-24\\ & =& 24\end{array}$

Proton to nucleon magnitude relation is;

$\begin{array}{rcl}\frac{\text{N}}{\text{Z}}& =& \frac{24}{24}\\ & =& 1:1\end{array}$

Proton tonucleonmagnitude relationof this nuclei isless thanthe zone of stability.

Therefore, it'll bear antilepton emission or negatron capture.

c)

${}_{25}{}^{50}\text{Mn}$

Let us find the mode of decay.
The number of protons is Z = 25.

The number of neutrons is:

$\begin{array}{rcl}\text{N}& =& \text{A}-\text{Z}\\ & =& 50-25\\ & =& 25\end{array}$

Proton to nucleon magnitude relation is

$\begin{array}{rcl}\frac{N}{Z}& =& \frac{25}{25}\\ & =& 1:1\end{array}$

Proton tonucleonmagnitude relationof this nuclei isless thanthe zone of stability.

Thereforeit'llbearantileptonemission ornegatroncapture.

The required graph is: