Question 65: Atoms & Nuclei

Question

List-I shows various functional dependencies of energy $(E)$ on the atomic number $(Z)$. Energies associated with certain phenomena are given in List-II.
Choose the option that describes the correct match between the entries in List-I to those in List-II.

List–I
List–II

(P) $E \propto Z^2$
(1) energy of characteristic x-rays

(Q) $E \propto (Z - 1)^2$
(2) electrostatic part of the nuclear binding energy for stable nuclei with mass numbers in the range 30 to 170

(R) $E \propto Z(Z - 1)$
(3) energy of continuous x-rays

(S) $E$ is practically independent of $Z$
(4) average nuclear binding energy per nucleon for stable nuclei with mass number in the range 30 to 170

(5) energy of radiation due to electronic transitions from hydrogen-like atoms

Accepted Answer

(P) For a hydrogen - like atom (atomic. number $z$ ), the energy of an
electron in the n-th orbit (Bohr model) is, $E_n=\frac{-13.6 z^2}{n^2} \mathrm{eV}$
So $E  \propto  z^2$
So energy of radiation due to electronic transitions from hydrogen like atoms can be written as
$$ \begin{aligned} & E=-13.6 z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}ight) \ \Rightarrow & E \propto z^2 \end{aligned} $$
$$ 	ext { so }(p) ightarrow(s) $$
(Q) Characteristic $x$-rays are emitted when an electron from a higher shell fills a vacancy in an inner shell (eg. K-shell). The energy of the emitted $x$-ray corresponds to the energy difference between the shells.
For $k$-shell $x$-rays,
$$ 	ext { L- shell }(n=2) ightarrow k 	ext {-shell }(n=1) $$ transition occurs.
The energy levels, are modified by screening effects as the inner electron sees an effective nuclear charge of approximately $(z-1)$ due to screening by the other $k$-shell electron.
So using Bohr-like model with effective charge :
$$ E_1=\frac{-13.6(z-1)^2}{1^2}, E_2=\frac{-13.6(z-1)^2}{2^2} $$
So $\Delta E=E_2-E_1$
$$ \begin{aligned} & =13 \cdot 6(z-1)^2\left[\frac{1}{1^2}-\frac{1}{2^2}ight] \ \Rightarrow & \Delta E \propto(z-1)^2 \ \Rightarrow & E \propto(z-1)^2 \end{aligned} $$
So $($ a $) \longrightarrow(1)$
(R) $E \propto z(z-1)$
Let's find the electrostatic energy between, protons in the nucleus: A nucleus with atomic number z has z protons.
The electrostatic (coulomb) energy arises from the repulsion between each pair of protons.
So number of paps $=z_{c_2}=\frac{z(z-1)}{2}$
Coulomb energy between two protons separated by a distance $r$.
$$ E_1=\frac{k q_1 q_2}{r}=\frac{1}{4 \pi \varepsilon_0} \frac{e \cdot e}{r}=\frac{e^2}{4 \pi \varepsilon_0 r} $$
So total electrostatic energy.
$$ \begin{aligned} E & =	ext { number of pairs } 	imes E_1 \ \Rightarrow E & =\frac{z(z-1)}{2} \cdot \frac{e^2}{4 \pi \varepsilon_0 r} \ \Rightarrow E & \propto z(z-1) \end{aligned} $$
So $(R) \longrightarrow(2)$
(s) The binding energy per nucleon is a measure of the stability of a nucleus and is given by the semi-empirical mass formula. For most nuclei (except very light ones), the average binding energy per nucleon is approximately constant, around $8-9 \mathrm{Mev}$ due to the saturation of nuclear forces. i.e., it is independent of $z$.
so ( S ) $ightarrow$ (4)
Hence, option (C) is correct.

Answer

P→4, Q→3, R→1, S→2

Answer

P→5, Q→2, R→1, S→4

Answer

P→5, Q→1, R→2, S→4

Answer

P→3, Q→2, R→1, S→5

Atoms & Nuclei

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