As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is single valence electron in Cr, the energy states may be thought of as given by the Bohr model.
The energy of the nth state
\(E_n=Z^2R{1\over n^2}\) where R is the Rydberg constant and Z = 24
In transition from n = 2 to n = 1,
Energy released
\(\Delta E=-RZ^2\left[{1\over4}-1\right]={3\over4}Z^2R\)
The energy required to eject a n = 4 electron is 
\(E_4=Z^2R{1\over16}={Z^2R\over16}\)
So K.E. of Auger electron is given by
K.E. \(=Z^2R\left({3\over4}-{1\over10}\right)\)
\(={11\over16}\times24\times24\times13.6eV\)
= 5385.6eV

\(Here,\ { r }_{ 0 }=2.8\times { 10 }^{ -14 }m,Z=79,\)
\( E=?\)
\( From\quad E=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \frac { (Ze)(2e) }{ { r }_{ 0 } } \)
\(E=\frac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \frac { { 2Ze }^{ 2 } }{ { r }_{ 0 } } =\frac { 9\times { 10 }^{ 9 }\times 2\times 79{ \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ 2.8\times { 10 }^{ -14 } }\)
\(=1.300\times { 10 }^{ -12 }J\)
\( =\frac { 1.300\times { 10 }^{ -12 } }{ 1.6\times { 10 }^{ -13 } } MeV=8.125MeV\)
\(As \ E \ is \ doubled,{ r }_{ 0 } \ becomes \ half\)
\( \frac { { r }_{ 0 } }{ 2 } =1.4\times { 10 }^{ -14 }m\)

(i) Nuclear reactors are used in electric power generation. They are also used to produce radioactive isotopes which have applications in medicine, industry and agriculture.
(ii) Radioactive waste consists of fission products and transuranic elements such as plutonium and americium. This waste is extremely hazardous to all forms of life on earth.
(iii) People often oppose the location site of a nuclear reactor because any leakage of nuclear radiations can affect adversely miles of area surrounding it. Elaborate safety measures are needed not only for reactor operations, but also for handling and disposal of radioactive waste.
I would suggest that Government must take stringent safety measures and assure people of safeguards in the event of nuclear accidents. At the same time, people must also be educated accordingly.

(i) Here,  \(N={ 10 }^{ 16 } \ and \ \frac { dN }{ dt } =4\times { 10 }^{ 16 } \ per \ sec.\lambda =?\)
\(From \ \frac { dN }{ dt } =\lambda N\)
\( \lambda =\frac { dN/dt }{ dt } =\frac { 4\times { 10 }^{ 16 } }{ { 10 }^{ 16 } } =4\times { 10 }^{ 16 }{ sec }^{ -1 }\)
(ii) The three types of radiactive are radiations are : \(\alpha \)-rays; \(\beta \)-rays and \(\gamma \)-rays. Out of then, \(\alpha \)-rays have maximum penetrating power.
(ii) Radioactive decay law is the law of mass action, i.e., larger the available mass, greater is its rate of decay. The same applies in day to day life. Larger the population, greater is their decay rate, i.e., death rate, i.e., under given conditions, when number of people is large, the rate at which they die is also large. The reverse is also true.

(i) Values shown here are dedication, hard working nature, brilliance, true justification of the help when helped by another, honesty and gratitude.
(ii) The phenomenon of spontaneous emission of  \(\alpha \) , \(\beta \) and \(\gamma \)-particles by the nuclide is known as radioactivity. Half-life period is related with the disintegration constant as,
\({ T }_{ 1/2 }=\frac { 0.693 }{ \lambda } \)
(iii) Average life of radioactive element,
\(\tau ={ 1.44T }_{ 1/2 }\)

(i) The total energy of electron in ground state (n = 1) is
\({ E }_{ 1 }=-\cfrac { 13.6 }{ { 1 }^{ 2 } } eV=-13.6eV\)          
It indicates that 13.6eV energy is needed to remove an electron from hydrogen atom in its ground state.
(ii) Energy in first excited state(n = 2) is
\({ E }_{ 2 }=-\cfrac { 13.6 }{ { 2 }^{ 2 } } eV=-3.4eV\)         
It implies that energy needed to remove an electron from hydrogen atom in first excited state is 3.4eV, which is less than 13.6eV.

(a) Definition
(i) Half life: Time taken by a radioactive nuclei to reduce to half of the initial number of radio nuclei.
(ii)Average life: Ratio of total life time of all radioactive nuclei to the total number of nuclei in the sample.
Relation between half life and decay constant:
\({ T }_{ 1/2 }=\frac { 0.693 }{ \lambda } \)
Relation between average life and decay constant
\(\tau =\frac { 1 }{ \lambda } \)
(b) \(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
\(\frac { 3 }{ 4 } { N }_{ 0 }={ N }_{ 0 }{ e }^{ -(0.3465)^{ t }\quad }\) = (\(​​​\because N\)75% of N₀)                      
\(N=\frac { 3 }{ 4 } { N }_{ 0 }\)
\({ e }^{ (0.3465)t }=\frac { 4 }{ 3 } \)
\(0.3465\times t={ log }_{ e }(4/3)\)
\(=2.303[log4-log3]\)
\(=2.303[0.6020-0.4771]\)
\( =2.303\times 0.1249\)
\( t=\frac { 2.303\times 0.1249 }{ 0.3465 } \)
\( \therefore \ t=0.83days\ or\ 19.92\ hours\)

(a) Keen observer/helpful/concerned/responsible/respectful towards elders(Any two)
(b) The doctor can trace and observe,the difference between the moment of an appropriate through a normal brain and a brain having tumour in it.

While drawing the plot. we have to keep in mind that first binding energy will increase sharply and then it will be constant almost.
For plot of binding energy per nucleon as the function of mass number A
Following are the two conclusions that can be drawn regarding the nature of the nuclear force.
The force is attractive and strong enough to produce a binding energy of few MeV per nucleon.
The two important conclusions regarding the nature of nuclear force are given below.
(i) The nuclear force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon.
(ii) The constancy of the binding energy in the wide range of mass number 30 < A < 170 indicate that nuclear force is a short-range force.
(b) (i) According to the binding energy curve, a very heavy nucleus (A > 170), has lower binding energy per nucleon compared to nuclei of middle mass number (30 < A < 170).
Thus, if a heavy nucleus breaks into two nuclei of mass number between 30 and 170, nucleons get more tightly bound. This implies energy would be released in the process. (nuclear fission)
(ii) When two light nuclei (A < 10) join to form a heavier nucleus, the binding energy per nucleon of fused heavier nucleus increases.
Again it indicates that energy would be released in the process (nuclear fusion).
(c) The basic nuclear process of neutron undergoing β-decay is given as
\(n \rightarrow p+e^{-}+\bar{v}\)
Here \(\bar{v}\) is antinutrino.
Neutrino and antineutrino both are neutral particles with very small (possibly, even zero) mass compared to the electrons. They have only weak interaction with other particles. Therefore, the detection of neutrinos is found very difficult.

It gives an estimate of the size of nucleus.
(ii) K.E of the ∝-particle = potential energy Possessed by beam at distance of closest approach.
\(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { (2e)(Ze) }{ { r }_{ 0 } } \)
7.7 x 1.6 x 10^-3 = \(\frac { 9\times { 10 }^{ 9 }\times 2\times 2.56\times { 10 }^{ -38 } }{ { r }_{ 0 } } \)
r₀ = \(\frac { 9\times { 10 }^{ 9 }\times 2\times 2.56\times { 10 }^{ -38 } }{ 7.7\times 1.6\times { 10 }^{ -13 } } m\)
= 299 x 10^-16 m