CBSE Class 12 Physics Modern Physics Exam Notes

CBSE Class 12 Physics Modern Physics Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

MODERN PHYSICS

Modern Physics – exam notes

BOHR’S MODEL OF THE ATOM

Bohr’s theory is a combination of classical and quantum approaches. It removes the drawbacks of Rutherford’s model, according to which an electron revolving around the nucleus should lose energy and finally fall into the nucleus. When an electron goes from one orbit to another, it can absorb or radiate electromagnetic radiation.

Postulates

1. The electron moves in a circular path around the nucleus under the influence of the electrostatic attraction between the electron and the nucleus.
2. The electron can move only in selected orbits in which its angular momentum is an integral multiple of \(\dfrac{h}{2\pi}\), i.e. \(mvr=\dfrac{nh}{2\pi}\), where \(h\) is Planck’s constant and \(n\) is the shell number.

Radii of the various orbits

Consider an electron of mass \(m\) and charge \(e\) revolving around the nucleus of charge \(Ze\) (\(Z\) = atomic number). Let \(v\) be the tangential velocity of the revolving electron and \(r\) the radius of the orbit. The electrostatic force of attraction between the nucleus and the electron is \(\dfrac{KZe^2}{r^2}\).

To move on a circular path,

\[\dfrac{KZe^2}{r^2}=\dfrac{mv^2}{r}\quad\text{or}\quad v^2=\dfrac{KZe^2}{rm}\qquad\cdots(i)\]

According to Bohr’s second postulate,

\[\text{angular momentum}=mvr=n\dfrac{h}{2\pi}\qquad\cdots(ii)\]

Solving (i) and (ii),

\[r_n=\dfrac{n^2h^2}{4\pi^2mKZe^2}=0.53\,\dfrac{n^2}{Z}\ \text{Å}\qquad\cdots(iii)\]

For the hydrogen atom, \(Z=1\), so \(r_n=\dfrac{n^2h^2}{4\pi^2mKe^2}=0.53\,n^2\) Å.

Velocity. Putting the value of \(r\) in equation (ii),

\[v_n=\dfrac{2\pi Ke^2}{nh}\,Z\quad\text{or}\quad v_n=2.18\times10^{6}\,\dfrac{Z}{n}\ \text{m/s}\]

Energy of an electron. The total energy of the electron is the sum of its kinetic and potential energies.

\[KE_n=\dfrac12mv_n^2=\dfrac{me^4Z^2}{8n^2h^2\epsilon_0^2}\]

\[PE_n=\dfrac{-1}{4\pi\epsilon_0}\cdot\dfrac{Ze^2}{r_n}=-\dfrac{me^4Z^2}{4n^2h^2\epsilon_0^2}\]

\[E_n=\text{T.E.}_n=KE_n+PE_n=-\dfrac{me^4Z^2}{8n^2h^2\epsilon_0^2}\]

Putting the values of the constants,

\[E_n=\dfrac{-13.6\,Z^2}{n^2}\ \text{eV}\]

The frequency of the revolving electron is

\[f=\dfrac{v_n}{2\pi r_n}=\dfrac{me^4Z^2}{4\epsilon_0^2n^3h^3}\]

INTERPRETATION OF THE HYDROGEN SPECTRUM

The electron in the hydrogen atom resides in the first orbit. When energy is supplied, the electron moves to higher energy shells depending on the amount of energy absorbed. When this electron returns to any of the lower energy levels, it emits energy. The Lyman series is obtained when the electron returns to the lowest energy state, while the Balmer series is obtained when the electron returns to the second energy shell. Similarly, the Paschen, Brackett and Pfund series are obtained when the electron returns to the third, fourth and fifth energy shell from higher energy shells respectively.

The maximum number of lines produced when an electron jumps from the \(n\)th level to the ground level is \(\dfrac{n(n-1)}{2}\). If the electron comes back from the energy level of energy \(E_2\) to the energy level of energy \(E_1\), the difference is expressed as the energy of a photon:

\[E_2-E_1=\Delta E=h\nu\quad\text{or}\quad \nu=\dfrac{\Delta E}{h}\]

Since \(\Delta E\) can have only definite values depending on the energies \(E_2\) and \(E_1\), \(\nu\) will have only fixed values in an atom.

\[\nu=\dfrac c\lambda=\dfrac{\Delta E}{h}\quad\text{or}\quad\lambda=\dfrac{hc}{\Delta E}\]

Since \(h\) and \(c\) are constants and \(\Delta E\) corresponds to a definite energy, each transition from one energy level to another produces light of a definite wavelength. This is actually observed as a line in the spectrum of the hydrogen atom. Thus the different spectral lines in the spectra of atoms correspond to different transitions of electrons from higher to lower energy levels.

Let \(n_1\) and \(n_2\) be the two energy shells in a hydrogen-like atom (\(n_1\lt n_2\)).

\[E_1=\dfrac{-13.6Z^2}{n_1^2}\ \text{eV},\qquad E_2=\dfrac{-13.6Z^2}{n_2^2}\ \text{eV}\]

\[E_2-E_1=13.6Z^2\left[\dfrac1{n_1^2}-\dfrac1{n_2^2}\right]\ \text{eV},\qquad \Delta E=\dfrac{hc}{\lambda}\]

\[\dfrac{hc}{\lambda}=13.6Z^2\left[\dfrac1{n_1^2}-\dfrac1{n_2^2}\right]\ \Rightarrow\ \dfrac1\lambda=RZ^2\left[\dfrac1{n_1^2}-\dfrac1{n_2^2}\right]\]

where \(R\) is the Rydberg constant for hydrogen. Its value is 109,679 cm\(^{-1}\).

List of series when the electron in a hydrogen atom goes from a higher orbit to a lower one:

The following points support Bohr’s theory:

1. The frequencies of the spectral lines calculated from Bohr’s equation are in agreement with the frequencies observed experimentally in the hydrogen spectrum.
2. The value of the Rydberg constant calculated from Bohr’s equation tallies with the value determined experimentally.
3. The spectra of hydrogen-like species such as He\(^+\), Li\(^{2+}\), Be\(^{3+}\) can be explained with the help of Bohr’s theory.

Limitations of Bohr’s theory

1. It does not explain the spectra of multi-electron atoms.
2. When a spectroscope of high resolving power is used, a spectral line in the hydrogen spectrum is found to be a collection of several lines very close to one another. Bohr’s theory does not explain this fine structure even for the hydrogen atom.
3. It does not explain the splitting of spectral lines into a group of lines under the influence of a magnetic field (Zeeman effect) and an electric field (Stark effect).
4. It is not in agreement with Heisenberg’s uncertainty principle.

X-RAYS

X-rays are electromagnetic radiations having wavelengths from a fraction of an angstrom (Å) to about 100 Å. They were discovered accidentally by Roentgen during his studies on electrical discharge in gases; he found that an unknown radiation was produced when electrons collided with the walls of the tube. The wave nature of X-rays was established by Laue, who demonstrated that they are diffracted by crystals.

Production. The modern X-ray tube, called the Coolidge tube, has a heated filament from which electrons are emitted. They are accelerated towards a cooled copper anode under a high potential difference (the cathode is concave so that it focuses the electrons). A target, which is a metal of high atomic number and high melting point, is embedded in the anode. The intensity of the X-ray beam is controlled by the filament current, which determines the number of electrons striking the target per unit time. The penetrating power (hardness) of the beam is controlled by the potential difference between the cathode and the anode.

X-ray spectra. A typical X-ray spectrum given by a target consists of a continuous range of wavelengths starting from a minimum value \(\lambda_{min}\). Superimposed on this is a characteristic line spectrum having sharp wavelengths.

Origin of the characteristic spectrum. If an incident electron collides with an electron in one of the inner shells and knocks it out of the atom, a vacancy is created in that shell. Another electron from a higher shell then makes a transition to this shell, creating another vacancy in the higher shell. The process of the jump of an electron from higher to lower shells continues till the inner shells are filled up. This generates a series of radiations, some of which lie in the X-ray region. These radiations are characteristic of the target element. The X-ray spectrum of a substance is classified into K-series, L-series, M-series, etc.

Moseley’s experiment and the concept of atomic number. Moseley studied the X-ray spectra of 38 different elements, from aluminium to gold. He measured the frequency of the principal lines of a particular series (the \(\alpha\)-lines in the K-series) of the spectra. He observed that the frequency of the particular spectral line was related to the serial number of the element in the periodic table, which he termed the atomic number (\(Z\)). He presented the relationship

\[\sqrt{\nu}=a(Z-b)\]

where \(\nu\) = frequency of the X-rays, \(Z\) = atomic number, and \(a\) and \(b\) are constants. When the square root of the frequency was plotted against the atomic number of the elements producing the X-rays, a straight line was obtained.

When an electron is knocked out from the \(n_1\) energy state and is filled by an electron from the \(n_2\) state, the wavelength of the X-ray emitted is given by

\[\dfrac1\lambda=R(Z-b)^2\left[\dfrac1{n_1^2}-\dfrac1{n_2^2}\right]\]

For the K-series, \(b=1\), \(n_1=1\) and \(n_2=2,3,\dots\) For the L-series, \(b=7.4\), \(n_1=2\) and \(n_2=3,4,\dots\)

For the K\(_\alpha\) line, the electron jumps from the L-level to a vacancy in the K-level. The L electron sees \(Z\) protons in the nucleus and one electron in the K-shell which screens the positive charge, so the net charge faced by the L electron can be taken as \((Z-1)e\). Now

\[E=Rhc(Z-1)^2\left[\dfrac1{1^2}-\dfrac1{2^2}\right]\]

\[h\nu=Rhc(Z-1)^2\left[\dfrac34\right]\]

\[\sqrt\nu=\sqrt{\dfrac34Rc}\,(Z-1)\]

Origin of the continuous spectrum. If an electron comes very close to a target nucleus, it is suddenly accelerated due to the Coulomb field of the nucleus and emits electromagnetic radiation. This radiation is called bremsstrahlung (braking radiation) and is continuous. The minimum wavelength (maximum frequency) corresponds to an electron losing all its energy in a single collision with a target atom. If \(V\) is the accelerating potential difference, then

\[h\nu_{max}=\dfrac{hc}{\lambda_{min}}=eV\]

\[\lambda_{min}=\dfrac{hc}{eV}=\dfrac{12420}{V}\ \text{Å}\]

Properties of X-rays

1. They ionise the material through which they pass.
2. They affect photographic plates in the same manner as visible light.
3. They cause fluorescence in certain chemical compounds like zinc sulphide.
4. They penetrate matter and are absorbed as they pass through it. If \(I_0\) is the intensity of the incident radiation and \(I\) is the intensity after travelling a distance \(x\), then \(I=I_0e^{-\mu x}\), where \(\mu\) is called the absorption coefficient of the material. The higher the atomic number of the material, the higher the absorption coefficient. This is the basis of radiography.
5. They cause photoelectric emission.
6. X-rays are not deflected by electric and magnetic fields, as they contain no charged particles.

Diffraction of X-rays by crystals. The spacing between atomic or molecular planes in crystals is of the order of a few angstroms. Since the wavelength of X-rays is also of this order, X-rays can be diffracted by crystals. If \(d\) is the spacing between the planes and \(\lambda\) is the wavelength of the X-rays, diffraction maxima are obtained at glancing angle \(\theta\) given by

\[2d\sin\theta=n\lambda\]

where \(n\) is an integer. This is called Bragg’s law.

Uses of X-rays. X-rays are used widely to detect and study bone fractures, diseases of the lungs and tooth decay. They are also used in cancer therapy, in industry and in materials-science research, and to detect structural defects, faults in joints, welding, etc.

PHOTOELECTRIC EFFECT

The emission of electrons from the surface of a metal when illuminated by electromagnetic radiation of suitable frequency is called the photoelectric effect. For a given material there exists a minimum frequency of incident radiation which emits photoelectrons. If the frequency is less than that, no electron can be emitted from the metal surface, no matter how intense the radiation may be. This minimum value of frequency is called the threshold frequency. The energy of the photoelectrons is proportional to the frequency of the incident radiation, and the number of photoelectrons emitted per second is proportional to the intensity of the incident radiation.

Accepting Planck’s idea that radiation consists of photons, Einstein suggested that the photon’s energy is used in two ways:

1. A part of the energy is used to free the electron from the atom. This energy is known as the photoelectric work function, denoted by \(W\), and is defined as the minimum energy required to emit electrons from the metal surface.
2. The rest of the energy is used in giving kinetic energy to the electron. Thus the energy of the photon is \(h\nu=W+\dfrac12mv^2\) …(i), where \(\nu\) is the frequency of the incident radiation and \(v\) is the velocity of the emitted electron.

When the photon’s energy is equal to the work function, the kinetic energy becomes zero. Under this condition, equation (i) reduces to

\[h\nu_0=W\]

where \(\nu_0\) is the threshold frequency. From equation (i),

\[\dfrac12mv^2=h\nu-W=h\nu-h\nu_0=h(\nu-\nu_0)\]

For a definite metal the value of \(h\nu_0\) is constant, so \(\dfrac12mv^2\) increases with \(h\nu\); that is, \(v^2\propto(\nu-\nu_0)\). Thus an increase of the frequency of the incident radiation above the threshold frequency causes an increase in the velocity of the electron.

WAVE NATURE OF ELECTRONS (DE BROGLIE’S HYPOTHESIS)

de Broglie proposed that an electron, like light, behaves both as a material particle and as a wave, and derived an expression for the wavelength of the wave associated with the electron. According to Planck’s equation

\[E=h\nu=h\dfrac c\lambda\qquad\cdots(i)\]

The energy of a photon on the basis of Einstein’s mass–energy relationship is

\[E=mc^2\qquad\cdots(ii)\]

where \(c\) is the velocity of the photon. Equating both equations,

\[h\dfrac c\lambda=mc^2\quad\text{or}\quad\lambda=\dfrac{h}{mc}\qquad\cdots(iii)\]

For a particle of mass \(m\) moving with velocity \(v\), the same relation gives \(\lambda=\dfrac{h}{mv}\). The wavelength decreases if the mass \(m\) increases; the de Broglie relation is applicable to small particles like the electron and has no practical significance for large particles.

ATOMIC NUCLEUS

Properties: the following are the properties of atomic nuclei.

1. Composition: all nuclei contain protons and neutrons except the ordinary hydrogen atom, which has only a single proton. A proton has charge \(+e\) and a neutron is neutral. Mass number of a nucleus \(A=Z+N\), where \(Z\) = number of protons in the nucleus and \(N\) = number of neutrons. A nucleus is represented symbolically as \(^A_ZX\).
2. Mass: nuclear mass has been measured accurately using a mass spectrometer. It is convenient to express mass in terms of the atomic mass unit (amu), defined as \(\dfrac1{12}\)th the mass of the carbon isotope \(^{12}_6C\). \(1\ \text{amu}=1.66\times10^{-27}\) kg. According to Einstein’s equation \(E=mc^2\), the energy equivalent of 1 amu is \(\dfrac{1.66\times10^{-27}\times(3\times10^8)^2}{1.6\times10^{-19}}\) eV \(=931\) MeV.
3. Size: most nuclei are almost spherical and the average radius can be expressed as \(R=R_0A^{1/3}\), where \(R_0\) is a constant called the nuclear unit radius.
4. Density: since the mass of a neutron is approximately the same as that of a proton, the mass of a nucleus can be written as \(mA\), where \(m\) is the mass of a proton. Then \(\rho=\dfrac MV=\dfrac{mA}{\frac43\pi(R_0A^{1/3})^3}=\dfrac{3m}{4\pi R_0^3}=\text{constant}=2.3\times10^{17}\ \text{kg/m}^3\). \(\rho\) is independent of \(A\), and is therefore the same for all nuclei.
5. Nuclei are of three types: (i) isotopes: nuclei having the same atomic number \(Z\) but different mass number \(A\), e.g. \(^1_1H,\ ^2_1H,\ ^3_1H\); (ii) isobars: nuclei having the same mass number \(A\) but different atomic number \(Z\), e.g. \(^{14}_6C\) and \(^{14}_7N\); (iii) isotones: nuclei having the same number of neutrons, e.g. \(^3_1H\) and \(^4_2He\).

RADIOACTIVITY

Becquerel, in 1896, discovered accidentally that uranium salt crystals emit an invisible radiation which affected a photographic plate even though it was properly covered. Further investigations by Marie and Pierre Curie and other workers showed that many other substances also emitted similar radiations. This property of spontaneous emission of radiation is called radioactivity. Subsequent work, notably of Rutherford, suggested that radioactivity was in fact due to the decay or disintegration of unstable nuclei.

Emission of \(\alpha\) particles: during \(\alpha\)-particle emission the atomic number reduces by 2 and the mass number reduces by 4, i.e. \(^A_ZX\to\,^{A-4}_{Z-2}Y+\,^4_2He\).

Emission of \(\beta\) particles: when a nucleus has excess neutrons, it emits a \(\beta\)-particle to bring the \(n/p\) ratio into the stable region. A neutron gets converted into a proton and a \(\beta\)-particle; therefore the mass number remains constant while the atomic number increases by 1.

\(\gamma\)-radiation: after emission of an \(\alpha\) or \(\beta\) particle the nucleus is left in an excited state. The nucleus comes to the stable state by emitting electromagnetic radiation known as \(\gamma\)-radiation. There is no change in \(A\) or \(Z\) during this process. \(\alpha\) and \(\beta\) emission do not take place simultaneously, while \(\gamma\)-radiation can be emitted along with either of them.

Properties of \(\alpha\), \(\beta\) and \(\gamma\) rays

Law of disintegration. It was studied by Rutherford and Soddy in 1902. The disintegration of nuclei is purely statistical, which means that nuclei take different times to disintegrate and each decays independently. The rate of disintegration is directly proportional to the number of undecayed nuclei present at that time, i.e.

\[-\dfrac{dN}{dt}\propto N=\lambda N\qquad\cdots(i)\]

where \(\lambda\) is the disintegration or decay constant. Integrating equation (i),

\[\log_eN=-\lambda t+C\]

At \(t=0\), \(N=N_0\), so \(C=\log_eN_0\). Hence \(\log_eN=-\lambda t+\log_eN_0\), or

\[N=N_0e^{-\lambda t}\qquad\cdots(ii)\]

Equation (ii) shows that the number of nuclei of a given radioactive substance decreases exponentially with time. It also shows that decay occurs rapidly at first and the rate of decay decreases with time.

Half-life (\(T_{1/2}\)): the time in which half of a radioactive substance decays. At \(t=T_{1/2}\), \(N=\dfrac{N_0}2\), so

\[\dfrac{N_0}{2}=N_0e^{-\lambda T_{1/2}}\ \Rightarrow\ T_{1/2}=\dfrac{\log_e2}{\lambda}=\dfrac{0.693}{\lambda}\qquad\cdots(iii)\]

If \(t=nT_{1/2}\), where \(n\) is an integer, equation (ii) reduces to \(N=N_0\left(\dfrac12\right)^n\).

Mean life (\(\tau\)): the mean life of a radioactive substance is defined as the sum of the lifetimes of all the radioactive nuclei divided by the total number of nuclei.

\[\tau=\dfrac{\int t\,dN}{\int dN}=\dfrac{\int t\,dN}{N_0}\quad\text{or}\quad\tau=\dfrac1\lambda\qquad\cdots(iv)\]

If \(t=\tau\), \(N=N_0e^{-\lambda(1/\lambda)}=0.37\,N_0\), i.e. in one mean life the radioactive substance decays by nearly 63%. From (iii) and (iv),

\[T_{1/2}=0.693\,\tau\qquad\cdots(v)\]

Units of rate of decay or activity. A number of units are used to express the activity of a radioactive sample. The more commonly used ones are:

1. Curie (Ci): the activity of a radioactive sample is one curie when \(3.7\times10^{10}\) decays take place per second, \(1\ \text{Ci}\equiv3.7\times10^{10}\) decays/s. This is the approximate activity of 1 g of radium. In practice the smaller units millicurie and microcurie are used: \(1\ \text{mCi}=3.7\times10^{7}\) decays/s and \(1\ \mu\text{Ci}=3.7\times10^{4}\) decays/s.
2. Becquerel (Bq): the SI unit of activity, representing 1 decay per second, \(1\ \text{Bq}\equiv1\) decay/s. We thus have \(1\ \text{Ci}=3.7\times10^{10}\) Bq.
3. Rutherford (Rd): another unit of activity, representing \(10^{6}\) decays per second, \(1\ \text{Rd}\equiv10^{6}\) decays/s.

Radioactive dating. Radioactive dating, also called carbon dating, is used to estimate the age of organic samples. The technique is based on the \(\beta\)-activity of the radio-isotope C\(^{14}\): \(^{14}_6C\to\,^{14}_7N+\beta^-+\bar\nu\). High-energy particles from outer space, called cosmic rays, induce nuclear reactions in the upper atmosphere and create carbon-14. The carbon dioxide molecules of the earth’s atmosphere have a constant ratio (\(\approx1.3\times10^{-12}\)) of C\(^{14}\) to C\(^{12}\). All living organisms also show the same ratio as they continuously exchange CO\(_2\) with their surroundings. However, after its death an organism can no longer absorb CO\(_2\) and the ratio C\(^{14}\)/C\(^{12}\) decreases due to the \(\beta\)-decay of C\(^{14}\). Thus by measuring the \(\beta\)-activity per unit mass, it is possible to estimate the age of a material. Using such techniques, samples of wood, charcoal, bone, etc., have been found to have lived from 1000 to 25000 years ago.

Mass defect. The actual mass of a nucleus (determined by a mass spectrometer of high resolving power) is always less than the sum of the masses of the protons and neutrons in the free state.

\[\Delta m=[Zm_p+(A-Z)m_n]-M\]

where \(m_p\) is the mass of a proton, \(m_n\) is the mass of a neutron and \(M\) is the mass of the nucleus.

Binding energy (B.E.): it is defined as the energy released during the formation of a nucleus as a result of the disappearance of mass, i.e. the mass defect.

\[\text{Binding energy}=(\Delta m)c^2,\qquad \text{binding energy per nucleon}=\dfrac{(\Delta m)c^2}{A}\]

Variation of B.E. per nucleon with mass number. If the average binding energy per nucleon is calculated for all nuclides and the results are plotted against the mass number \(A\), the graph shown is obtained.

It is observed from the graph that the binding energy per nucleon (except for He\(^4\), C\(^{12}\) and O\(^{16}\)) rises first sharply and reaches a maximum value of 8.8 MeV in the neighbourhood of \(A=50\). The curve then falls very slowly, reaching 8.4 MeV at about \(A=140\). For higher mass numbers the energy decreases to about 7.6 MeV.

NUCLEAR FISSION

Fission was first observed by the German scientists Otto Hahn and Fritz Strassmann in 1938. In nuclear fission a heavy nucleus splits into two smaller nuclei with the liberation of energy. When uranium (\(Z=92\)) is bombarded with neutrons, it splits into two fragments, namely barium (\(Z=56\)) and krypton (\(Z=36\)), and a large amount of energy is released, which appears due to the decrease in mass. The reaction is represented as

\[^{235}_{92}U+\,^1_0n\to\,^{141}_{56}Ba+\,^{92}_{36}Kr+3\,^1_0n+\text{energy}\]

The disintegration process in which a heavy nucleus, after capturing a neutron, splits up into nuclei of nearly equal mass is called nuclear fission.

Energy released in nuclear fission. The amount of energy released in nuclear fission may be obtained by the method of mass defect. For example, consider the fission of uranium-235 (\(Z=92\)) into barium-141 (\(Z=56\)) and krypton-92 (\(Z=36\)) by slow neutrons:

\[^{235}_{92}U+\,^1_0n\to\,^{236}_{92}U\to\,^{141}_{56}Ba+\,^{92}_{36}Kr+3\,^1_0n+Q\]

Mass decrease during the nuclear reaction \(=236.133-235.910=0.223\) amu, so the corresponding energy release is \(0.223\times931\approx200\) MeV (about 208 MeV by direct multiplication).

If we calculate the energy produced by one gram of uranium, it comes out to be \(2.28\times10^{4}\) kWh \(=22.8\) MWh. This shows that 1 kg of uranium-235 would give a power of 1 MW for more than two years.

NUCLEAR FUSION

The binding energy versus \(A\) graph shows that when lighter nuclei with \(A\lt20\) combine to form bigger nuclei, the binding energy per nucleon increases. The total binding energy of the product is greater than the total binding energy of the reactants, resulting in the release of energy. This process of combining two lighter nuclei into a bigger one is known as nuclear fusion, i.e.

\[^2_1H+\,^2_1H\to\,^4_2He+24\ \text{MeV}\]

The following points deserve particular attention concerning nuclear fusion.

1. The energy of 21.62 MeV released in one fusion event is much smaller than the about 200 MeV released in one fission event. This does not mean that fusion is a weaker source of energy than fission. If we compare the energy released per unit mass, one fusion event is accompanied by a release of \(\dfrac{21.62\ \text{MeV}}{6\ \text{amu}}\) or 3.6 MeV per amu, as against \(\dfrac{200\ \text{MeV}}{235}\) or 0.85 MeV per amu released in one fission event.
2. For fusion, positively charged nuclei have to come very close to each other. This requires a very high energy to be provided to the fusing nuclei to enable them to overcome the strong electrostatic repulsion between them. Calculations show that the necessary energy can be provided by raising the temperature to about \(10^8\) K. Such a high temperature can be produced by first inducing a fission event. A fusion reaction is therefore also called a thermonuclear reaction. This is the basis of the hydrogen bomb.
3. Unlike the highly radioactive fission fragments, the end product of the fusion of hydrogen nuclei is safe, non-radioactive helium.
4. Unfortunately, a sustained and controllable fusion reactor that can deliver a net power output is not yet a reality. A great deal of effort is currently under way to resolve various difficulties in the development of a successful device. Nevertheless, controlled fusion is regarded as the ultimate energy source because of the abundant availability of its main fuel, water.

Interpretation of hydrogen spectrum

Modern Physics MCQ Questions with Answers Class 12 Physics