Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately $10^{-15} \mathrm{~m}$.
Each particle (neutron and proton) present inside the nucleus is called a nucleon.
Let $\lambda$ be the wavelength $\lambda=10^{-15} \mathrm{~m}$
To detect separate parts inside a nucleon, the electron must have wavelength less than $10^{-15} \mathrm{~m}$.
We know that
$$\begin{aligned} \lambda & =\frac{h}{p} \text { and } K E=P E \quad\text{.... (i)}\\ \text{Energy}\quad & =\frac{h c}{\lambda}\quad\text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { From Eq. (i) and Eq. (ii), }\\ &\begin{aligned} \text { kinetic energy of electron } & =\mathrm{PE}=\frac{h \mathrm{c}}{\lambda}-\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ \mathrm{KE} & =10^9 \mathrm{eV} \end{aligned} \end{aligned}$$
A nuclide 1 is said to be the mirror isobar of nuclide 2 if $Z_1=N_2$ and $Z_2=N_1$.(a) What nuclide is a mirror isobar of ${ }_{11}^{23} \mathrm{Na}$ ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?
(a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2 , if $Z_1=N_2$ and $Z_2=N_1$.
Now in ${ }_{11} \mathrm{Na}^{23}, Z_1=11, N_1=23-11=12$
$\therefore$ Mirror isobar of ${ }_{11} \mathrm{Na}^{23}$ is ${ }_{12} \mathrm{Mg}^{23}$, for which $Z_2=12=N_1$ and $N_2=23-12=11=Z_1$
(b) $\mathrm{As}_{12}^{23} \mathrm{Mg}$ contains even number of protons (12) against ${ }_{11}^{23} \mathrm{Na}$ which has odd number of protons (11), therefore ${ }_{11}^{23} \mathrm{Mg}$ has greater binding energy than ${ }_{11} \mathrm{Na}^{23}$.
Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
$${ }^{38} \text { Sulphur } \xrightarrow[=2.48 \mathrm{~h}]{\text { half-life }}{ }^{38} \mathrm{Cl} \xrightarrow[=0.62 \mathrm{~h}]{\text { half-life }}{ }^{38} \mathrm{Ar} \text { (stable) }$$
Assume that we start with $1000{ }^{38} \mathrm{~S}$ nuclei at time $t=0$. The number of ${ }^{38} \mathrm{Cl}$ is of count zero at $t=0$ and will again be zero at $t=\infty$. At what value of $t$, would the number of counts be a maximum?
Consider the chain of two decays
$${ }^{38} \mathrm{~S} \xrightarrow[2.48 \mathrm{~h}]{ }{ }^{38} \mathrm{Cl} \xrightarrow[0.62 \mathrm{~h}]{ }{ }^{38} \mathrm{Ar}$$
At time $t$, Let ${ }^{38} \mathrm{~S}$ have $N_1(t)$ active nuclei and ${ }^{38} \mathrm{Cl}$ have $N_2(t)$ active nuclei.
$$\begin{aligned} \frac{d N_1}{d t} & =-\lambda_1 N_1=\text { rate of formation of } \mathrm{Cl}^{38} \\ \text{Also,}\quad\frac{d N_2}{d t} & =-\lambda_1 N_2+\lambda_1 N_1 \\ \text{But}\quad N_1 & =N_0 e^{-\lambda_1 t} \\ \frac{d N_2}{d t} & =\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2\quad\text{.... (i)} \end{aligned}$$
Multiplying by $e^{\lambda z^t} d t$ and rearranging
$$e^{\lambda_2 t} d N_2+\lambda_2 N_2 e^{\lambda_2 t} d t=\lambda_1 N_0 e^{\left(\lambda_2-\lambda_1\right) t} d t$$
Integrating both sides
$$N_2 e^{\lambda_2 t}=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1} e^{\left(\lambda_2-\lambda_1\right) t}+C$$
$$\begin{aligned} \therefore \quad N_2 e^{\lambda_2 t} & =\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}\left(e^{\left(\lambda_2-\lambda_1\right) t}-1\right) \quad\text{.... (ii)}\\ N_2 & =\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right) \end{aligned}$$
For maximum count, $\quad \frac{d N_2}{d t}=0$
$$\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2=0\quad$$ [From Eq. (i)]
$\Rightarrow \quad \frac{N_0}{N_2}=\frac{\lambda_2}{\lambda_1} e^{\lambda_1 t}\quad$ [From Eq. (ii)]
$$\begin{gathered} e^{\lambda_2 t}-\frac{\lambda_2}{\lambda_1} \cdot \frac{\lambda_1}{\left(\lambda_1-\lambda_1\right)} e^{\lambda_1 t}\left[e^{\left(\lambda_2-\lambda_1\right) t}-1\right]=0 \\ \text{or}\quad e^{\lambda_2 t}-\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\lambda_2 t}+\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\lambda_1 t}=0 \end{gathered}$$
$$\begin{aligned} 1-\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)}+\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\left(\lambda_1-\lambda_2\right) t} & =0 \\ \frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\left(\lambda_1-\lambda_2\right) t} & =\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)}-1 \\ e^{\left(\lambda_1-\lambda_2\right) t} & =\frac{\lambda_1}{\lambda_2} \\ t & =\left(\log _e \frac{\lambda_1}{\lambda_2}\right) /\left(\lambda_1-\lambda_2\right) \\ & =\frac{\log _e\left(\frac{2.48}{0.62}\right)}{2.48-0.62} \\ & =\frac{\log _e 4}{1.86}=\frac{2.303 \times 2 \times 0.3010}{1.86} \quad\left(\because \lambda=\frac{0.693}{T_{1 / 2}}\right) \\ & =0.745 \mathrm{~S} \end{aligned}$$
Deuteron is a bound state of a neutron and a proton with a binding energy $B=2.2 \mathrm{MeV}$. A $\gamma$-ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma$-ray. If $E=B$, show that this cannot happen. Hence, calculate how much bigger than $B$ must be $E$ be for such a process to happen.
Given binding energy $\quad B=2.2 \mathrm{~MeV}$
From the energy conservation law,
$$E-B=K_n+K_p=\frac{p_n^2}{2 m}+\frac{p_p^2}{2 m}\quad\text{.... (i)}$$
$$\begin{aligned} &\text { From conservation of momentum, }\\ &\begin{array}{ll} & p_n+p_p=\frac{E}{C} \quad\text{.... (ii)}\\ \text { As } & E=B, \text { Eq. (i) } p_n^2+p_p^2=0 \end{array} \end{aligned}$$
It only happen if $p_n=p_p=0$
So, the Eq. (ii) cannot satisfied and the process cannot take place.
Let $E=B+X$, where $X< Put value of $p_{\mathrm{n}}$ from Eq. (ii) in Eq. (i), we get $$\begin{aligned}
X & =\frac{\left(\frac{E}{c}-p_{\mathrm{p}}\right)}{2 m}+\frac{p_{\mathrm{p}}^2}{2 m} \\
\text { or } \quad 2 p_{\mathrm{p}}^2-\frac{2 E p_{\mathrm{p}}}{c}+\frac{E^2}{c^2}-2 m X & =0
\end{aligned}$$ Using the formula of quadratic equation, we get $$p_p=\frac{\frac{2 E}{c} \pm \sqrt{\frac{4 E^2}{c^2}-8\left(\frac{E^2}{c^2}-2 m X\right)}}{4}$$ For the real value $p_p$, the discriminant is positive $$\begin{aligned}
\frac{4 E^2}{c^2} & =8\left(\frac{E^2}{c^2}-2 m X\right) \\
16 m X & =\frac{4 E^2}{c^2} \\
X & =\frac{E^2}{4 m c^2} \approx \frac{B^2}{4 m c^2} \quad[\therefore X<< B \Rightarrow E \cong B]
\end{aligned}$$
The deuteron is bound by nuclear forces just as H -atom is made up of $p$ and $e$ bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form a coulomb potential but with an effective charge $e^{\prime}$
$$F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^{\prime 2}}{r}$$
estimate the value of $\left(e^{\prime} / e\right)$ given that the binding energy of a deuteron is 2.2 MeV .
The binding energy is H -atom
$$E=\frac{m e^4}{\pi \varepsilon_0^2 h^2}=13.6 \mathrm{eV}\quad\text{.... (i)}$$
If proton and neutron had charge $e^{\prime}$ each and were governed by the same electrostatic force, then in the above equation we would need to replace electronic mass $m$ by the reduced mass $m^{\prime}$ of proton-neutron and the electronic charge e by $e^{\prime}$.
$$\begin{aligned} m^{\prime} & =\frac{M \times N}{M+N}=\frac{M}{2} \\ & =\frac{1836 \mathrm{~m}}{2}=918 \mathrm{~m} \end{aligned}$$
$$\begin{aligned} &\text { Here, } M \text { represents mass of a neutron/proton }\\ &\therefore \quad \text { Binding energy }=\frac{918 m\left(e^{\prime}\right)^4}{8 \varepsilon_0^2 h^2}=2.2 \mathrm{MeV}\quad\text{.... (ii)} \end{aligned}$$
$$\begin{aligned} &\text { Dividing Eqs. (ii) and (i), we get }\\ &\begin{aligned} 918\left(\frac{e^{\prime}}{e}\right)^4 & =\frac{2.2 \mathrm{MeV}}{13.6 \mathrm{eV}}=\frac{2.2 \times 10^6}{13.6} \\ \left(\frac{e^{\prime}}{e}\right)^4 & =\frac{2.2 \times 10^6}{13.6 \times 918}=176.21 \\ \frac{e^{\prime}}{e} & =(176.21)^{1 / 4}=3.64 \end{aligned} \end{aligned}$$