A plot of the number of neutrons (N) against | vedclass

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(C) For stable nuclei with $Z > 20$,the $N/Z$ ratio is greater than $1$. For an unstable nucleus with an $N/Z$ ratio less than $1$,the nucleus is prot

Vedclass · Accepted Answer

$B, D$

Vedclass · Answer

(C) For stable nuclei with $Z > 20$,the $N/Z$ ratio is greater than $1$. For an unstable nucleus with an $N/Z$ ratio less than $1$,the nucleus is proton-rich.To increase the $N/Z$ ratio and move towards the stability line,the nucleus must decrease the number of protons or increase the number of neutrons.$1$. $\beta^{+}$-decay (positron emission): $^A_Z X ightarrow ^A_{Z-1} Y + ^0_{+1} e + 
u_e$. Here,a proton is converted into a neutron,increasing the $N/Z$ ratio.$2$. $K$-electron capture: $^A_Z X + ^0_{-1} e ightarrow ^A_{Z-1} Y + u_e$. Here,an orbital electron is captured by the nucleus,converting a proton into a neutron,which also increases the $N/Z$ ratio.Both processes effectively increase the $N/Z$ ratio towards $1$,thereby stabilizing the nucleus. Thus,the correct modes are $B$ and $D$.

Similar Questions

If $M(A, Z)$,$M_p$,and $M_n$ denote the masses of the nucleus ${}_Z^AX$,proton,and neutron respectively in units of $u$ $(1u = 931.5 \text{ MeV}/c^2)$,and $BE$ represents its binding energy in $\text{MeV}$,then:

The mass of a nucleus $_Z^AX$ is denoted by $M(A, Z)$. If $M_p$ and $M_n$ are the masses of a proton and a neutron respectively, the binding energy of this nucleus is given by:

The binding energy of a nucleus is equivalent to

One atomic mass unit is equivalent to .............. $MeV$ energy.

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