I. Multiple Choice Questions:<\/span><\/p>\nQuestion 1.
\nSuppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is
\na) 14.4 \\(\\frac{Z}{V}\\) \u00c5
\nb) 14.4 \\(\\frac{\\mathrm{V}}{\\mathrm{Z}}\\) \u00c5
\nc) 1.44 \\(\\frac{Z}{V}\\) \u00c5
\nd) 1.44 \\(\\frac{\\mathrm{V}}{\\mathrm{Z}}\\) \u00c5
\nAnswer:
\nc) 1.44 \\(\\frac{Z}{V}\\) \u00c5
\nSolution:
\n<\/p>\n
Question 2.
\nIn a hydrogen atom, the electron revolving in the fourth orbit, has angular momentum equal to
\na) h
\nb) \\(\\frac{\\mathrm{h}}{\\pi}\\)
\nc) \\(\\frac{4 h}{\\pi}\\)
\nd) \\(\\frac{2 \\mathrm{~h}}{\\pi}\\)
\nAnswer:
\nd) \\(\\frac{2 \\mathrm{~h}}{\\pi}\\)
\nSolution:
\n<\/p>\n
Question 3.
\nAtomic number of H – like atom with ionization potential 122.4 V for n = 1 is
\na) 1
\nb) 2
\nc) 3
\nd) 4
\nAnswer:
\nc) 3
\nSolution:
\n<\/p>\n
Question 4.
\nThe ratio between the first three orbits of hydrogen atom is
\na) 1 : 2 : 3
\nb) 2 : 4 : 6
\nc) 1 : 4 : 9
\nd) 1 : 3 : 5
\nAnswer:
\nc) 1 : 4 : 9
\nSolution:
\nrn<\/sub> \u03b1 n2<\/sup>
\nr1<\/sub> : r2<\/sub> : r3<\/sub> = 1 : 4 : 9<\/p>\nQuestion 5.
\nThe charge of cathode rays is
\na) positive
\nb) negative
\nc) neutral
\nd) not defined
\nAnswer:
\nb) negative
\n(They are negatively charged particles)<\/p>\n
<\/p>\n
Question 6.
\nIn J.J. Thomson e\/m experiment, a beam of the electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
\n(a) B is increased by 208 times
\n(b) B is decreased by 208 times
\n(c) B is increased by 14.4 times
\n(d) B is decreased by 14.4 times
\nAnswer:
\n(c) B is increased by 14.4 times
\nHint:
\nIn the condition of no deflection \\(\\frac { e }{ m }\\) = \\(\\frac {{ E }^{2}}{{ 2vB }^{2}}\\)
\nIf m is increased by 208 times then B should be increased \\(\\sqrt { 208 } \\) = 14.4 time<\/p>\n
Question 7.
\nThe ratio of the wavelengths for the transition from n = 2 to n = 1 in Li++<\/sup>, He+<\/sup> and H is
\na) 1 : 2 : 3
\nb) 1 : 4 : 9
\nc) 3 : 2 : 1
\nd) 4 : 9 : 36
\nAnswer:
\nd) 4 : 9 : 36
\nSolution:
\n<\/p>\nQuestion 8.
\nThe electric potential between a proton and an electron is given by V = V0<\/sub> ln \\(\\left(\\frac{\\mathbf{r}}{\\mathbf{r}_{0}}\\right)\\) where r0<\/sub> is a constant. Assume that Bohr atom model is applicable to potential, then variation of radius nth orbit rn<\/sub> with the principal quantum number n is
\na) rn<\/sub> \u03b1 \\(\\frac{1}{\\mathrm{n}}\\)
\nb) rn<\/sub> \u03b1 n
\nc) rn<\/sub> \u03b1 \\(\\frac{1}{n^{2}}\\)
\nd) rn<\/sub> \u03b1 n2<\/sup>
\nAnswer:
\nb) rn<\/sub> \u03b1 n
\nSolution:
\n<\/p>\nQuestion 9.
\nIf the nuclear radius of 27<\/sup> Al is 3.6 fermi, the approximate unclear radius of64 <\/sup>Cu is
\n(a) 2.4
\n(b) 1.2
\n(c) 4.8
\n(d) 3.6
\nAnswer:
\n(c) 4.8
\nHint:
\n\\(\\frac {{ R }_{Al}}{{ R }_{Cu}}\\) = \\(\\frac{(27)^{1 \/ 3}}{(64)^{1 \/ 3}}\\) = \\(\\frac { 3 }{ 4}\\)
\nRcu<\/sub> = \\(\\frac { 4 }{ 3}\\) RAl <\/sub>= \\(\\frac { 4 }{ 3}\\) x 3.6 fermi
\nRcu<\/sub> = 4.8 fermi<\/p>\nQuestion 10.
\nThe nucleus is approximately spherical in shape. Then the surface area of the nucleus having mass number A varies as
\na) A2\/3<\/sup>
\nb) A4\/3<\/sup>
\nc) A1\/3<\/sup>
\nd) A5\/3<\/sup>
\nAnswer:
\na) A2\/3<\/sup>
\nSolution:
\nSurface area = 4\u03c0R\u00b2
\nSurface area \u03b1 R\u00b2
\nR \u03b1 A1\/3<\/sup>
\nSurface area \u03b1 (A1\/3<\/sup>)2<\/sup>
\nSurface area \u03b1 A2\/3<\/sup>
\n<\/p>\nQuestion 11.
\nThe mass of a \\({ }_{3}^{7} \\mathbf{L i}\\) nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of \\({ }_{3}^{7} \\mathbf{L i}\\) nucleus is nearly
\na) 46 MeV
\nb) 5.6 MeV
\nc) 3.9 MeV
\nd) 23 MeV
\nAnswer:
\nb) 5.6 MeV
\nSolution:
\n<\/p>\n
Question 12.
\nMp<\/sub> denotes the mass of the proton and Mn<\/sub> denotes the mass of a neutron. A given nucleus of binding energy B, contains Z protons and N neutrons. The mass M (N, Z) of the nucleus is given by(where c is the speed of light)
\n(a) M (N, Z) = NMn<\/sub> + ZMp<\/sub> – Bc2<\/sup>
\n(b) M (N, Z) = NMn<\/sub> + ZMp<\/sub> + Bc2<\/sup>
\n(c) M (N, Z) = NMn<\/sub> + ZMp<\/sub> – B \/ c2<\/sup>
\n(d) M (N, Z) = NMn<\/sub> + ZMp<\/sub> + B \/ c2<\/sup>
\nAnswer:
\n(c) M (N, Z) = NMn<\/sub> + ZMp<\/sub> – B \/ c2<\/sup>
\nHint:
\nBinding energy, B = [ZMp<\/sub> + NMn<\/sub> – M (N, Z)] C2<\/sup>
\nM(N,Z) = ZMp<\/sub> + NMn<\/sub> – \\(\\frac { B }{{ C }^{ 2 }}\\)<\/p>\nQuestion 13.
\nA radioactive nucleus (initial mass number A and atomic number Z) emits 2\u03b1 and 2 positrons. The ratio of number of neutrons to that of proton in the final nucleus will be
\na) \\(\\frac{A-Z-4}{Z-2}\\)<\/p>\n
b) \\(\\frac{A-Z-2}{Z-6}\\)<\/p>\n
c) \\(\\frac{A-Z-4}{Z-6}\\)<\/p>\n
d) \\(\\frac{A-Z-12}{Z-4}\\)
\nAnswer:
\nb) \\(\\frac{A-Z-2}{Z-6}\\)<\/p>\n
Question 14.
\nThe half-life period of radioactive element A is same as the mean lifetime of another radioactive element B. Initially both have the same number of atoms. Then.
\na) A and B have the same decay rate initially
\nb) A and B decay at the same rate always
\nc) B will decay at faster rate than A
\nd) A will decay at faster rate than B.
\nAnswer:
\nc) B will decay at faster rate than A
\nSolution:
\n(t1\/2<\/sub>)A<\/sub> = (tmean<\/sub>)B<\/sub>
\n\\(\\frac{0.693}{\\lambda_{\\mathrm{A}}}\\) = \\(\\frac{1}{\\lambda_{\\mathrm{B}}}\\)
\n\u03bbA<\/sub> = 0.693\u03bbB<\/sub>
\n\u03bbA<\/sub> < \u03bbB<\/sub>
\nrate decay = \u03bbN
\nIntially no. of atoms (N) of both are equal but since \u03bbB<\/sub> > \u03bbA<\/sub>, B will decay at a faster
\nrate than A.<\/p>\nQuestion 15.
\nA system consists of N0<\/sub> nucleus at t = 0. The number of nuclei remaining after half of a half – life (that is, at time t = \\(\\frac{1}{2}\\)T1\/2<\/sub>)
\na) \\(\\frac{\\mathrm{N}_{0}}{2}\\)<\/p>\nb) \\(\\frac{N_{0}}{\\sqrt{2}}\\)<\/p>\n
c) \\(\\frac{\\mathrm{N}_{0}}{4}\\)<\/p>\n
d) \\(\\frac{\\mathrm{N}_{0}}{8}\\)
\nAnswer:
\n(b) \\(\\frac{N_{0}}{\\sqrt{2}}\\)) N0
\nSolution:
\n<\/p>\n
<\/p>\n
II. Short answer questions:<\/span><\/p>\nQuestion 1.
\nWhat are cathode rays?
\nAnswer:
\nA cathode ray is a stream of electrons that are seen in vaccum tubes. It is called a “cathode ray” because the electrons are being emitted from the negative charged element in the vaccum tube called the cathode.<\/p>\n
Question 2.
\nWrite the properties of cathode rays.
\nAnswer:
\nProperties of cathode rays:<\/p>\n
\n- Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 107<\/sup> ms-1<\/sup>. It can be deflected by the application of electric and magnetic fields. The direction of deflection indicates that they are negatively charged particles.<\/li>\n
- When the cathode rays are allowed to fall of matter, they produce heat. They affect the photographic plates and also produce fluorescence when they fall on certain crystals and minerals.<\/li>\n
- When the cathode rays fall on a material of high atomic weight, x-rays are produced.<\/li>\n
- Cathode rays ionize the gas through which they pass.<\/li>\n
- The speed of rys is upto \\(\\left(\\frac{1}{10}\\right)^{\\text {th }}\\) of the speed of light.<\/li>\n<\/ol>\n
Question 3.
\nGive the results of Rutherford alpha scattering experiment.
\nAnswer:<\/p>\n
\n- Most of the alpha particles are un\u00acdeflected through the gold foil and went straight.<\/li>\n
- Some of the alpha particles are deflected through a small angle.<\/li>\n
- A few alpha particles (one in thousand) are deflected through an angle more than 90\u00b0<\/li>\n<\/ol>\n
Question 4.
\nWrite down the postulates of Bohr atom model.
\nAnswer:
\n1. The electron in an atom moves around nucleus in circular orbits under the influence of Coulomb electrostatic force of attraction. This Coulomb force gives necessary centripetal force for the electron to undergo circular motion.
\n2. Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy. Only those discrete orbits allowed are stable orbits.
\n3. The angular momentum of the electrons in these stationary orbits are quantized that is, it can be written as integer or integral multiple of \\(\\frac{\\mathrm{h}}{2 \\mathrm{~h}}\\) called as reduced Planck’s constant – that is, h (read it as h-bar) and the integer n is called as principal quantum number of the orbit.
\nl = nh;
\nL = \\(\\frac{\\mathrm{nh}}{2 \\pi}\\);
\nmvr = \\(\\frac{\\mathrm{nh}}{2 \\pi}\\) where h = \\(\\frac{\\mathrm{h}}{2 \\pi}\\)<\/p>\n
This condition is known as angular momentum quantization condition.
\n4. An electron can jump from one orbit to another orbit by absorbing or emitting a photon whose energy is equal to the difference in energy (\u2206E) between the two orbital levels.
\nenergy quantization condition
\n\u2206E = Efinal<\/sub> – Einitial<\/sub> = h\u03bd = \\(\\frac{\\mathrm{hc}}{\\lambda}\\)
\n\u03bb – Wavelength or radiation
\nC – Speed of light
\n\u03bd – Frequency of the radiation<\/p>\n<\/p>\n
Question 5.
\nWhat is meant by excitation energy?
\nAnswer:
\nExcitation energy and excitation potential:
\nThe energy required to excite an electron from lower energy state to any higher energy state is known as excitation energy.<\/p>\n
Question 6.
\nDefine the ionization ionization potential.
\nAnswer:
\nThe minimum energy required to remove an electron from an atom in the ground state is known as binding energy or ionization energy.
\nEionization<\/sub> = E\u221e<\/sub> – En<\/sub>
\n= 0 – (-\\(\\frac{13.6}{\\mathrm{n}^{2}}\\) Z2<\/sup>eV)
\n= \\(\\frac{13.6}{\\mathrm{n}^{2}}\\) Z2<\/sup>eV
\nIonization potential is defined as ionization energy per unit charge.
\nVionization<\/sub> = \\(\\frac{1}{\\mathrm{e}}\\) Eionization<\/sub> = \\(\\frac{13.6}{\\mathrm{n}^{2}}\\) Z2<\/sup>V<\/p>\nQuestion 7.
\nWrite down the drawbacks of Bohr atom model.
\nAnswer:
\nLimitations of Bohr atom model:
\nThe following are the drawbacks of Bohr atom model:<\/p>\n
\n- Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for complex atoms.<\/li>\n
- When the spectral lines are closely examined, individual lines of hydrogen spectrum is accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.<\/li>\n
- Bohr atom model fails to explain the intensity variations in the spectral lines.<\/li>\n
- The distribution of electrons in atoms is not completely explained by Bohr atom model.<\/li>\n<\/ol>\n
Question 8.
\nWhat is distance of closest approach?
\nAnswer:
\nThe minimum distance between the centre of the nucleus and the alpha particle just before it gets reflected back through 180\u00b0 is defined as the distance of closest approach r0<\/sub> (also known as contact distance).
\n<\/p>\nQuestion 9.
\nDefine impact parameter.
\nAnswer:
\nThe impact parameter is defined as the perpendicular distance between the centre of the gold nucleus and the direction of velocity vector of alpha particle when it is at a large distance.<\/p>\n
Question 10.
\nWrite a general notation of the nucleus of element X. What each term denotes?
\nAnswer:
\nGeneral notation of nucleus of element X.
\nZ<\/sub>a<\/sup>X
\nwhere x is the chemical symbol of the element
\nA is the mass number and Z is the atomic number.
\nFor example \\(7^{\\mathrm{N}^{15}}\\)
\nZ = 7, N = (A – Z) = (15 – 7) = 8<\/p>\n<\/p>\n
Question 11.
\nWhat is isotope? Give an example.
\nAnswer:
\nIsotopes are atoms of the same element having same atomic number Z, but different mass number A. For example, hydrogen has three isotopes and they are represented as \\({ }_{1}^{1} \\mathrm{H}\\) {H (hydrogen), \\({ }_{1}^{2} \\mathrm{H}\\) (deuterium), and \\({ }_{1}^{3} \\mathrm{H}\\) (tritium). Note that all the three nuclei have one proton and, hydrogen has no neutron, deuterium has 1 neutron and tritium has 2 neutrons.<\/p>\n
Question 12.
\nWhat is isotone? Give an example.
\nAnswer:
\nIsotones are the atoms of different elements having same number of neutrons. \\(_{ 5 }^{ 12 }B\\) and \\(_{ 6 }^{ 13 }B\\) are examples of isotones which 7 neutrons.<\/p>\n
Question 13.
\nWhat is isobar? Give an example.
\nAnswer:
\nIsobars are the atoms of different elements having the same mass number A, but different atomic number Z.
\nFor example \\({ }_{16}^{40} \\mathrm{~S}, \\stackrel{40}{17} \\mathrm{Cl}, \\stackrel{40}{18} \\mathrm{Ar}, \\underset{19}{40} \\mathrm{~K}\\) and \\({ }_{20}^{40} \\mathrm{Ca}\\) are isobars having same mass number 40 and different atomic number. In other words, isobars are the atoms of different chemical element which has same number of nucleon.<\/p>\n
Question 14.
\nDefine atomic mass unit u.
\nAnswer:
\nOne atomic mass unit (u) is defined as the 1\/12th<\/sup> of the mass of the isotope of carbon \\(_{ 6 }^{ 12 }C\\).<\/p>\nQuestion 15.
\nShow that nuclear density is almost constant for nuclei with Z > 10.
\nAnswer:
\n
\nThe nuclear density is independent of the mass number A. The nuclear density is almost constant for all the nuclei (Z > 10) irrespective of its size.<\/p>\n
<\/p>\n
Question 16.
\nWhat is mass defect?
\nAnswer:
\nIn general, if M, mp and m are mass of the nucleus (\\({ }_{Z}^{A} \\mathrm{X}\\)), the mass of proton and the mass of neutron respectively
\nmass defect \u2206m = (Zmp<\/sub> + Nmn<\/sub> ) \u2014 M difference in mass \u2206m is called mass defect.
\nThe mass of any nucleus is always less than the sum of the mass of its indivitual constituents. The difference in mass \u2206m is called mass defect.<\/p>\nQuestion 17.
\nWhat is binding energy of a nucleus? Give its expression.
\nAnswer:
\nwhen Z protons and N neutrons combine to form a nucleus, mass equal to mass defect disappears and the corresponding energy is released. This is called the binding energy of the nucleus (BE) and is equal to (\u2206m) c2<\/sup>.
\nBE = (Zmp<\/sub> + Nmn<\/sub> – M ) c2<\/sup><\/p>\nQuestion 18.
\nCalculate the energy equivalent of 1 atomic mass unit.
\nAnswer:
\nUsing Einstein’s mass-energy equivalence, the energy equivalent of one atomic mass unit
\n1u = 1.66 \u00d7 10-27<\/sup> \u00d7 (3 \u00d7 108<\/sup>)2<\/sup>
\n= 14.94 \u00d7 10-11<\/sup> J \u2248 931 MeV<\/p>\nQuestion 19.
\nGive the physical meaning of binding energy per nucleon.
\nAnswer:
\nThe average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus.
\n\\(\\overline{\\mathrm{BE}}\\) = \\(\\frac{\\left[\\mathrm{Zm}_{\\mathrm{H}}+\\mathrm{Nm}_{\\mathrm{n}}-\\mathrm{M}_{\\mathrm{A}}\\right] \\mathrm{C}^{2}}{\\mathrm{~A}}\\)<\/p>\n
Question 20.
\nWhat is meant by radioactivity?
\nAnswer:
\nThe phenomenon of spontaneous emission of highly penetrating radiations such as \u03b1, \u03b2 and \u03b3 rays by an element is called radioactivity.<\/p>\n
<\/p>\n
Question 21.
\nGive the symbolic representation of alpha decay, beta decay and gamma decay.
\nAnswer:
\nAlpha decay:
\n\\({ }_{Z}^{A} \\mathrm{X}\\) \u2192 \\(\\begin{array}{l}
\n\\mathrm{A}-4 \\\\
\n\\mathrm{Z}-2
\n\\end{array} \\mathrm{Y}\\) + \\({ }_{2}^{4} \\mathrm{He}\\)
\nHere X is called the parent nucleus and Y is called the daughter nucleus.<\/p>\n
\u03b2–<\/sup> decay:
\n\\({ }_{Z}^{A} \\mathrm{X}\\) \u2192 \\(\\begin{array}{c}
\n\\mathrm{A} \\\\
\n\\mathrm{Z}+1
\n\\end{array} \\mathrm{Y}\\) + e–<\/sup> + \\(\\bar{v}\\)
\nIt implies that the element X becomes Y by giving out an electron and antineutrino \\(\\bar{V}\\)<\/p>\n\u03b2+<\/sup> decay:
\n\\({ }_{Z}^{A} \\mathrm{X}\\) \u2192 \\(\\begin{array}{c}
\n\\mathrm{A} \\\\
\n\\mathrm{Z}-1
\n\\end{array} \\mathrm{Y}\\) + e+<\/sup> + \u03bd
\nIt implies that the element X becomes Y by giving out an positron and neutrino (v).<\/p>\nGamma decay:
\n\\({ }_{Z}^{A} \\mathrm{X*}\\) \u2192 \\({ }_{Z}^{A} \\mathrm{Y}\\) + gamma (\u03b3) rays
\nHere the asterisk (*) means excited state nucleus. In gamma decay, there is no change in the mass number or atomic number of the nucleus.<\/p>\n
Question 22.
\nIn alpha decay, why the unstable nucleus emits \\({ }_{2}^{4} \\mathrm{He}\\) nucleus? Why it does not emit four separate nucleons?
\nAnswer:
\nAfter all \\({ }_{2}^{4} \\mathrm{He}\\) consists of two protons and two neutrons. For example, if \\({ }_{92}^{238} \\mathrm{U}\\) nucleus decays into \\({ }_{90}^{234} \\mathrm{Th}\\) by emitting four separate nucleons (two protons and two neutrons), then the disintegration energy Q for this process turns out to be negative. It implies that the total mass of products is greater than that of parent \\({ }_{92}^{238} \\mathrm{U}\\) nucleus. This kind of process cannot occur in nature because it would Violate conservation of energy. In any decay process, the conservation of energy, conservation of linear momentum and conservation of angular momentum must be obeyed.<\/p>\n
Question 23.
\nWhat is mean life of nucleus? Give the expression.
\nAnswer:
\nThe mean lifetime of the nucleus is the ratio of sum or integration of life times of all nuclei to the total number nuclei present initially. The expression for mean life time, \u03c4 = \\(\\frac { 1 }{ \u03bb }\\).
\n\u03c4 = \\(\\frac{1}{\\lambda}\\)<\/p>\n
Question 24.
\nWhat is half – life of nucleus? Give the expression.
\nAnswer:
\nThe half – life T1\/2<\/sub> is the time required for the number of atoms initally present to reduce to one half of the initial amount.
\nT1\/2<\/sub> = \\(\\frac{\\ln 2}{\\lambda}\\) = \\(\\frac{0.6931}{\\lambda}\\)<\/p>\n<\/p>\n
Question 25.
\nWhat is meant by activity (or) decay rate? Give its unit.
\nAnswer:
\nActivity (R) or decay rate which is the number of nuclei decayed per second and it is denoted as
\nR \\(=\\left|\\frac{\\mathrm{dN}}{\\mathrm{dt}}\\right|\\).<\/p>\n
R = \\(\\left|\\frac{\\mathrm{dN}}{\\mathrm{dt}}\\right|\\) = \u03bbN0<\/sub>e-\u03bbt<\/sup><\/p>\nR = R0<\/sub>e-\u03bbt<\/sup>
\nThe SI unit of activity R is Becquerel and one Becquerel (Bq) is equal to one decay per second. There is also another standard unit for the activity called Curie (Ci).
\n1 Curie = 1 Ci = 3.7 \u00d7 1010<\/sup> decays per second
\n1 Ci = 3.7 \u00d7 1010<\/sup> Bq<\/p>\nQuestion 26.
\nDefine curie.
\nAnswer:
\nOne curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 1010<\/sup> decays\/s.<\/p>\nQuestion 27.
\nWhat are the constituent particles of neutron and proton?
\nAnswer:
\nAccording to quark model, proton is made up of two up quarks and one down quark and neutron is made up of one up quark and two down quark.<\/p>\n
<\/p>\n
III. Long answer questions:<\/span><\/p>\nQuestion 1.
\nExplain the J.J. Thomson experiment to determine the specific charge of electron.
\nAnswer:
\n1. In 1887, J.J. Thomson, measured the specific charge (e\/m) of electron.
\n2. The specific charge is defined as the charge per unit mass of particle.
\nPrinciple:
\n3. In the presence of electric and magnetic fields, the cathode rays are deflected.<\/p>\n
\nArrangement of J.J. Thomson experiment to determine the specific charge of an electron<\/p>\n
4. A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A.
\n5. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays.
\n6. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage.
\n7. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to eacth other.
\n8. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed.
\n9. This is achieved by coating the screen with zinc sulphide.<\/p>\n
(i) Determination of velocity of cathode rays :
\n1. For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O
\n2. i.e magnitude of electric force is balanced by the magnitude of force due to magnetic field.
\neE = eB\u03bd
\n=>\u03bd = \\(\\frac{E}{B}\\) ………………(1)
\ne- charge of the cathode rays
\n3. Electric force balancing the magnetic force – the path of electron beam is a straight line.<\/p>\n
ii) Determination of specific charge:
\n1. Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode.
\n2. Let V be the potential difference between anode and cathode, then the potential energy is eV.
\n3. Then from law of conservation of energy.
\ne V = \\(\\frac{1}{2}\\) m\u03bd2<\/sup>
\n\u21d2 \\(\\frac{e}{m}=\\frac{1}{2 V}\\) \u03bd2<\/sup>
\n4. Substituting the value of velocity from equation (1), we get
\n\\(\\frac{\\mathrm{e}}{\\mathrm{m}}=\\frac{1}{2 \\mathrm{~V}} \\frac{\\mathrm{E}^{2}}{\\mathrm{~B}^{2}}\\)
\n5. Substituting the values of E, B and V, the specific charge can be determined as
\n\\(\\frac{\\mathrm{e}}{\\mathrm{m}}\\) = 1.7 \u00d7 1011<\/sup> C kg-1<\/sup><\/p>\n<\/p>\n
Question 2.
\nDiscuss the Millikan\u2019s oil drop experiment to determine the charge of an electron.
\nAnswer:
\nThe Millikan’s experiment is used to determine one of the fundamental constants of nature known as charge of an electron.<\/p>\n
Principle:
\n1. This method is based on the study of the motion of uncharged oil drop under free fall due to gravity and charged oil drop in a uniform electric field.
\n2. By adjusting electric field suitably, the motion of oil drop inside the chamber can be controlled – that is, it can be made to move up or down or even kept balanced in the field of view for sufficiently long time.<\/p>\n
Experimental arrangement:
\n1. The apparatus consists of two horizontal circular metal plates A and B each with diameter around 20 cm and are separated by a small distance 1.5 cm.<\/p>\n
\nMillkan’s Experiment<\/p>\n
2. These two parallel plates are enclosed in a chamber with glass walls. Further, plates A and B are given a high potential difference around 10 kV such that electric field acts vertically downward.
\n3. A small hole is made at the centre of the upper plate A and atomizer is kept exactly above the hole to spray the liquid.
\n4. When a fine droplet of highly viscous liquid (like glycerine) is sprayed using atomizer, it falls freely downward through the hole of the top plate only under the influence of gravity.
\n5. Few oil drops in the chamber can acquire electric charge (negative charge) because of friction with air or passage of x – rays in between the parallel plates.
\n6. Further the chamber is illuminated by light which is passed horizontally and oil drops can be seen clearly using microscope placed perpendicular to the light beam. These drops can move either upwards or downward.<\/p>\n
(a) Determination of radius of the droplet:
\nFg<\/sub> = gravitational force
\nFb<\/sub> = buoyant force
\nFv<\/sub> = viscous force
\nviscous force and buoyant force balance the gravitational force.
\n7. When the electric field is switched off, the oil drop accelerates downwards. Due to the presence of air drag forces, the oil drops easily attain its terminal velocity and moves with constant velocity.<\/p>\n
\n(a) Free body diagram of the oil drop without electric field
\n(b) Free body diagram of the oil drop with electric field<\/p>\n
8. Let the gravitational force acting on the oil drop (downward) be Fb<\/sub> = mg, m = mass of the oil drop g = Acceleration due to gravity.
\n9. Let \u03c1 be the density of the oil drop, and r be the radius of the oil drop, then the mass of the oil drop can be expressed in terms of its density as
\n\u03c1 = \\(\\frac{\\mathrm{m}}{\\mathrm{V}}\\)
\nm = \u03c1(\\(\\frac{4}{3}\\) \u03c0r3)<\/sup>\u00a0(volume of the sphere, V = \\(\\frac{4}{3}\\) \u03c0r3<\/sup>)
\n10. The gravitational force can be written in terms of density as
\nFg<\/sub> = mg
\n=>Fg<\/sub> = \u03c1(\\(\\frac{4}{3}\\)) \u03c0r3<\/sup>
\n11. Let \u03c3 be the density of the air, the upthrust force experienced by the oil drop due to displaced air is
\nFb<\/sub> = \u03c3(\\(\\frac{4}{3}\\) \u03c0r3<\/sup>)g
\n12. From Stokes law, the viscous force on the oil drop is.
\nFv<\/sub> = 6\u03c0r\u03c5\u03b7
\nThe force balancing equation is
\nFg<\/sub> = Fb<\/sub> + Fv<\/sub>
\n\u03c1(\\(\\frac{4}{3}\\) \u03c0r3<\/sup>)g = \u03c3(\\(\\frac{4}{3}\\) \u03c0r3<\/sup>)g + 6\u03c0r\u03c5\u03b7
\n\\(\\frac{4}{3}\\) \u03c0r3<\/sup>(\u03c1 – \u03c3) = 6\u03c0r\u03c5\u03b7<\/p>\n\\(\\frac{2}{3}\\) \u03c0r3<\/sup>(\u03c1 – \u03c3) = 3\u03c0r\u03c5\u03b7<\/p>\nr = \\(\\left[\\frac{9 \\eta v}{2(\\rho-\\sigma) g}\\right]^{\\frac{1}{2}}\\) …………….(1)
\nThus, equation (1) gives the radius of the oil drop.<\/p>\n
(b) Determination of electric charge:
\n(a) Fe = qE
\n(c) buoyant force Fb<\/sub>
\n13. When the electric field is switched on, charged oil drops experience an upward electric force (qE).
\n14. Among many drops, one particulars drop can be chosen in the field of view of microscope and strength of the electric field is adjusted to make that particular drop to be stationary.
\n15. Under these circumstances, there will be no viscous force acting on the oil drop. Force acting on the oil droplet is
\nFe<\/sub> + Fb<\/sub> = Fg<\/sub>
\n\u21d2 qE + \\(\\frac{4}{3}\\) \u03c0r3<\/sup> \u03c3g = \\(\\frac{4}{3}\\) \u03c0r3<\/sup> \u03c1g
\n\u21d2 qE = \\(\\frac{4}{3}\\) \u03c0r3<\/sup> (\u03c1 – \u03c3)g
\n\u21d2 q = \\(\\frac{4}{3 \\mathrm{E}}\\) \u03c0r3<\/sup> (\u03c1 – \u03c3)g …………..(2)<\/p>\nSubstituting equation (1) in equation (2), we get
\nq = \\(\\frac{18 \\pi}{\\mathrm{E}}\\left(\\frac{\\eta^{3} v^{3}}{2(\\rho-\\sigma) \\mathrm{g}}\\right)^{\\frac{1}{2}}\\)
\n16. Millikan repeated this experiment several times and computed the charges on oil drops. He found that the charge of any oil drop can be written as integral multiple of a basic value, (e = 1.6 \u00d7 10-19<\/sup> C), which is nothing but the charge of an electron.<\/p>\n<\/p>\n
Question 3.
\nDerive the energy expression for hydrogen atom using Bohr atom model.
\nAnswer:
\nEnergy of a Hydrogen atom:
\nBohr postulates is find the allowed energies of the atom for different allowed orbits of the electron.
\n(a) The electron in an atom moves around nucleus in circular orbits under the influence of coulomb electrostatic force of attraction.
\n(b) Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy.
\nAngular momentum quantization condition,
\nL = n\\(\\hbar\\) = \\(\\frac{\\mathrm{nh}}{2 \\pi}\\)
\nh – planck’s constant
\nn – principal quantum number of the orbit.
\nc) An electron can jump from one orbit to another orbit by absorbing or emitting a photon whose energy is equal to the difference in energy (\u2206E) between the two orbital levels.
\nenergy quantization condition
\n\u2206E = Efinal<\/sub> – Einitial<\/sub> = h\u03c5 = \\(\\frac{\\mathrm{hc}}{\\lambda}\\)
\n\u03bb – wavelength of the radiation
\nc – speed of light
\n\u03c5 – frequency of the radiation<\/p>\nRadius of the orbit of the electron:<\/p>\n
\nElectron revolving around the nucleus<\/p>\n
1. The nucleus has a positive charge +Ze.
\n2. Let z be the atomic number of the atom.
\n3. Let -e be the charge of the electron.
\n4. The nucleus at rest and an electron revolving around the nucleus in a circular orbit of radius rn<\/sub> with a constant speed \u03c5n<\/sub>
\nCoulomb’s law, the force of attraction between the nucleus and the electron is
\n\\(\\overrightarrow{\\mathrm{F}}_{\\text {coulomb }}\\) = \\(\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{(+\\mathrm{Ze})(-\\mathrm{e})}{\\mathrm{r}_{\\mathrm{n}}^{2}} \\hat{\\mathrm{r}}\\)<\/p>\n= –\\(\\frac{1}{4 \\pi \\varepsilon_{0}} \\frac{Z \\mathrm{e}^{2}}{\\mathrm{r}_{\\mathrm{n}}^{2}} \\hat{\\mathrm{r}}\\)<\/p>\n
5. This force provides necessary centripetal force.
\n\\(\\overrightarrow{\\mathrm{F}}_{\\text {centripetal }}=\\frac{m v_{n}^{2}}{r_{n}} \\hat{r}\\)<\/p>\n
where m be the mass of the electron that moves with a velocity \u03c5n<\/sub> in a circular orbit.
\nTherfore,
\n
\nWhere n\u2208N \u03b50<\/sub>, n, e and \u03c0 \u2192 are constant
\nradius of the orbit,
\nrn<\/sub> = a0<\/sub> \\(\\frac{n^{2}}{Z}\\)<\/p>\nBohr radius a0<\/sub> = \\(\\frac{\\varepsilon_{0} h^{2}}{\\pi m e^{2}}\\)
\na0<\/sub> = 0.529 \u00c5
\nBohr radius is also used as unit of length called Bohr
\n1 Bohr = 0.53 \u00c5
\nFor hydrogen atom Z = 1
\nrn<\/sub> = a0<\/sub> n2<\/sup>
\nrn<\/sub> \u03b1 n2<\/sup>
\nFor the first orbit (ground sate)
\nr1<\/sub> = a0<\/sub>
\nr2<\/sub> = 4a0<\/sub> = 4r1<\/sub>
\nr3<\/sub> = 9a0<\/sub> = 9r1<\/sub>
\nr4<\/sub> = 16a0<\/sub> = 16r1<\/sub><\/p>\nThe energy of an electron in the nth orbit:
\nSince the electrostatic force is a conservative force, the potential energy for the nth orbit is
\n
\n6. The negative sign in the equation indicates that the electron is bound to the nucleus
\n7. where n stands for a principal quantum number.
\n8. The energies of the excited states come closer and closer together when the principal quantum number n takes higher values.
\n9. The ground state energy of hydrogen (-13.6 eV) is used as a unit of energy called Rydberg (lRydberg = -13.6eV).
\n10. For the first orbit (ground state), the total energy of electron is E1<\/sub> = -13.6 eV.
\n11. For the second orbit (first excited state), the total energy of electron is E2<\/sub> = – 3.4eV.
\n12. For the third orbit (second excited state), the total energy of electron ie E3<\/sub> = -1.51 eV and so on.
\n13. When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.<\/p>\n<\/p>\n
Question 4.
\nDiscuss the spectral series of hydrogen atom.
\nAnswer:
\nSpectral series of hydrogen atom:
\n1. As electron in excited states have very small lifetime, these electrons jump back to ground state through spontaneous emission in a short duration of time (approximately 10) by emitting the radiation with same wavelength (or frequency) corresponding to the colours is absorbed. This is called emission spectroscopy.<\/p>\n
<\/p>\n
\nSpectral series – Lyman, Balmer, Paschen series<\/p>\n
2. In each series, the distance of separation between the consecutive wavelengths decreases from higher wavelength to the lower wavelength, and also wavelength in each series approach a limiting value known as the series limit.
\n3. The wavelengths of these spectral lines perfectly agree with the equaion derived from Bohr atom model.
\n\\(\\frac{1}{\\lambda}=\\mathrm{R}\\left(\\frac{1}{\\mathrm{n}^{2}}-\\frac{1}{\\mathrm{~m}^{2}}\\right)=\\overline{\\mathrm{v}}\\) …………..(1)<\/p>\n
where \\(\\bar{v}\\) is known as wave number which is inverse of wavelength, R is known as Rydberg
\nconstant whose value is 1.09737 \u00d7 107<\/sup> m-1<\/sup> and m and n are positive integers such that m > n.
\nThe various spectral series are discussed below:
\n<\/p>\nQuestion 5.
\nExplain the variation of average binding energy with the mass number by graph and discuss its features.
\nAnswer:
\nBinding energy curve :
\n1. \\(\\overline{\\mathrm{BE}}\\) is plotted against A of all known nuclei.
\n2. It gives a curve as seen in Figure.
\n3. average binding energy per nucleon \\(\\overline{\\mathrm{BE}}\\).
\nIt is given by,<\/p>\n\\(\\overline{\\mathrm{BE}}=\\frac{\\left[\\mathrm{Zm}_{\\mathrm{H}}+\\mathrm{Nm}_{\\mathrm{n}}-\\mathrm{M}_{\\mathrm{A}}\\right] \\mathrm{c}^{2}}{\\mathrm{~A}}\\)\n
\nAvg. binding energy of the nucleus<\/p>\n
4. The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus.<\/p>\n
Important inferences from of the average binding energy curve:
\n1. The value of \\(\\overline{\\mathrm{BE}}\\) rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (iron) and then it slowly decreases.
\n2. The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
\n3. For higher mass numbers, the curve reduces slowly and \\(\\overline{\\mathrm{BE}}\\) for uranium is about 7.6 MeV. They are unstable and radioactive.
\nFrom Figure if two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
\n4. If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.<\/p>\n
<\/p>\n
Question 6.
\nExplain in detail the nuclear force.
\nAnswer:
\n1. Nucleus contains protons and neutrons. From electrostatics, like charges repel each other.
\n2. In the nucleus, the protons are separated by a distance of about a few Fermi (10-15<\/sup> m), they must exert on each other a very strong repulsive force. For example,
\n<\/p>\n3. This is nearly 1028<\/sup> times greater than the acceleration due to gravity.
\n4. From this observation, it was concluded that there must be a strong attractive force between protons to overcome the repulsive Coulomb\u2019s force.
\n5. This attractive force which holds the nucleus together is called strong nuclear force.
\n6. A few properties of strong nuclear force are the strong nuclear force is of very short range, acting only up to a distance of a few Fermi.
\n7. But inside the nucleus, the repulsive Coulomb force or attractive gravitational forces between two protons are much weaker than the strong nuclear force between two protons.
\n8. Similarly, the gravitational force between two neutrons is also much weaker than strong nuclear force between the neutrons. So nuclear force is the strongest force in nature.
\n9. The strong nuclear force is attractive and acts with an equal strength between proton-proton, proton-neutron. and neutron – neutron.
\n10. Strong nuclear force does not act on the electrons. So it does not alter the chemical properties of the atom.<\/p>\nQuestion 7.
\nDiscuss the alpha decay process with an example.
\nAnswer:
\nAlpha decay:
\n1. When unstable nuclei decay by emitting an a – particle (\\({ }_{2}^{4} \\mathrm{He}\\) nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. The alpha decay process symbolically represented as.
\n\\({ }_{Z}^{A} \\mathrm{X}\\) \u2192 \\({ }_{Z-2}^{A-4} \\mathrm{Y}\\) + \\({ }_{2}^{4} \\mathrm{He}\\)<\/p>\n
2. Here X is called the parent nucleus and Y is called the daughter nucleus.
\n3. Example:
\nDecay of Uranium \\({ }_{92}^{238} \\mathrm{U}\\) to thorium \\({ }_{90}^{234} \\mathrm{Th}\\) with the emission of \\({ }_{2}^{4} \\mathrm{He}\\) nucleus (\u03b1 – particle)<\/p>\n
\\({ }_{92}^{238} \\mathrm{U}\\) \u2192 \\({ }_{90}^{234} \\mathrm{Th}\\) + \\({ }_{2}^{4} \\mathrm{He}\\)<\/p>\n
4. As already mentioned, the total mass of the daughter nucleus and \\({ }_{2}^{4} \\mathrm{He}\\) nucleus is always
\n5. less than that of the parent nucleus. The difference in mass (\u2206m = mx<\/sub> – my<\/sub> -m\u03b1<\/sub> ) is released as energy called disintegration energy Q and is given by
\nQ = (mx<\/sub> – my<\/sub> – m\u03b1<\/sub>) c2<\/sup>
\n6. For spontaneous decay (natural radioactivity) Q > 0. In alpha decay process, the disintegration energy is certainly positive (Q > 0).
\n7. In fact, the disintegration energy Q is also the net kinetic energy gained in the decay process or if the parent nucleus is at rest, Q is the total kinetic energy of daughter nucleus and the \\({ }_{2}^{4} \\mathrm{He}\\) nucleus.
\n8. Suppose Q < 0, then the decay process cannot occur spontaneously and energy must be supplied to induce the decay.<\/p>\n<\/p>\n
Question 8.
\nDiscuss the beta decay process with examples.
\nAnswer:<\/p>\n
In beta decay, a radioactive nucleus emits either electron or positron. If electron (e–<\/sup>) is emitted, it is called \u03b2–<\/sup> decay and if positron (e+<\/sup>) is emitted, it is called p+ decay. The positron is an anti-particle of an electron whose mass is same as that of electron and charge is opposite to that of electron – that is, +e. Both positron and electron are referred to as beta particles.<\/p>\n1. \u03b2–<\/sup> decay:
\nIn \u03b2–<\/sup> decay, the atomic number of the nucleus increases by one but mass number remains the same. This decay is represented by
\n\\(_{ Z }^{ A }{ X }\\) \u2192 \\(_{ Z+12 }^{ A }{ Y}\\) + e–<\/sup> + \\(\\bar { v } \\) …(1)
\nIt implies that the element X becomes Y by giving out an electron and antineutrino (\\(\\bar { v } \\)). In otherwords, in each \u03b2–<\/sup> decay, one neutron in the nucleus of X is converted into a proton by emitting an electron (e–<\/sup>) and antineutrino. It is given by
\nn \u2192 p + e–<\/sup> + \\(\\bar { v } \\)
\nWhere p -proton, \\(\\bar { v } \\) -antineutrino. Example: Carbon (\\(_{ 6 }^{ 14 }{ C }\\)) is converted into nitrogen (\\(_{ 7 }^{ 14 }{ N }\\)) through \u03b2- decay.
\n\\(_{ 6 }^{ 14 }{ C }\\) \u2192 \\(_{ 7 }^{ 14 }{ N }\\) + e–<\/sup> + \\(\\bar { v } \\)<\/p>\n2. \u03b2+<\/sup> decay:
\nIn p+ decay, the atomic number is decreased by one and the mass number remains the same. This decay is represented by
\n\\(_{ Z }^{ A }{ X }\\) \u2192 \\(_{ Z-12 }^{ A }{ Y}\\) + e+<\/sup> + v
\nIt implies that the element X becomes Y by giving out a positron and neutrino (v). In other words, for each \u03b2+<\/sup> decay, a proton in the nucleus of X is converted into a neutron by emitting a positron (e+<\/sup>) and a neutrino. It is given by
\np \u2192 n + e+<\/sup> + v<\/p>\nHowever, a single proton (not inside any nucleus) cannot have \u03b2+<\/sup> decay due to energy conservation, because neutron mass is larger than proton mass. But a single neutron (not inside any nucleus) can have \u03b2–<\/sup> decay.
\nExample: Sodium (\\(_{ 11 }^{ 23 }{ Na }\\)) is converted into neon (\\(_{ 10 }^{ 22 }{ Ne }\\)) decay.
\n\\(_{ 11 }^{ 23 }{ Na }\\) \u2192 \\(_{ 10 }^{ 22 }{ Ne }\\) + e+<\/sup> + v<\/p>\nBeta-decay:
\nIn beta decay, a radioactive nucleus emits either electron or positron. If electron (e–<\/sup>) is emitted, it is called \u03b2