Sure shot questions for class XII CBSE Board Exam reg. 02(Two) Marks Questions and Answer

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1 ATOMS AND NUCLEI Sure shot questions for class XII CBSE Board Exam reg. 02(Two) Marks Questions and Answer Q.1. Define the term activity of a radio nuclide & draw the graph showing the variation of decay rate with number of active nuclei. Ans.:- Activity of a radionuclide is defined as the rate of disintegration of given radioactive sample with time. Its S.I. Unit is becquerel. We know that = N Where is constant for a given radioactive material so the graph between &N is straight line as shown in diagraph given below. dn/dt N Q.2. Theenergy of the electron in the ground state of hydrogen atom is e V (n=1). i. What does the negative sign signify? ii. How much energy is required to take an electron in this atom from the ground state to the first excited state? Ans.i. Negative sign signifies that the electron is bound to the nucleus. ii.energy in ground state = ev Energy in 1 st excited state = 13.6/2 2 = -3.4 ev Energy required to carry the electron to the first excited state = -3.4 (-13.6) = = 10.2 ev Q.3. Write two characteristic features of nuclear force which distinguish it from coulomb s force? Ans.:- Characteristic features of nuclear force i. Nuclear forces are short range attractive forces (range 2 to 3 fm) while coulomb s forces are of infinite range may be attractive or repulsive. ii. Nuclear forces are charge independent forces, while coulomb s force acts only between charged particles.

2 03 Marks Questions and Answer Q.4. Draw a graph of radioactive species versus time. Two different radioactive elements with half-life T 1 & T 2 have N 1 and N 2 (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant. Ans. The required graph is shown in the diagram. We know,activity R (= ) = N & decay constant = log e 2/T Therefore, Activity R = log e 2 N/T Therefore, R 1 = log e 2 N 1 /T 1, & R 2 = log e 2 N 2 /T 2 Therefore, R 1 /R 2 = N 1 T 2 /N 2 T 1

Q.5. Draw a graph showing the variation of binding energy per nucleon with the mass number. What are man inferences from the graph? Ans. Graph:- Main Inference from the graph i.

3 Q.5. Draw a graph showing the variation of binding energy per nucleon with the mass number. What are man inferences from the graph? Ans. Graph:- Main Inference from the graph i. The nuclei having mass number below 20and above 180 have relatively small binding energy per nucleon a hence they are unstable. ii. The nuclei having mass number about 56 have maximum binding energy per nucleon 8.5 Mev& hence they are stable. iii. Some nuclei have peaks (ex He 4, 2 C 12 6,O 16 8 ) this indicates that those nuclei are relatively more stable than their neighbors. Q.6. Aradioactive isotope has a half life of 5 years after how much time is its activity reduce to 3.125% of its original activity? Ans. We know = ) n Given = 3.125% = = ) n Or, = ) n Or, ) 5 n = 5 ) n We Know n = t = nt = 5*5 years ( T 5 years) = 25 years

05 Marks Questions&Answers Q.7. Describe Geiger- Marsden (Rutherford) experiment. Give its observations and conclusions? Ans. At the suggestion, Rutherford In 1911, H.

4 05 Marks Questions&Answers Q.7. Describe Geiger- Marsden (Rutherford) experiment. Give its observations and conclusions? Ans. At the suggestion, Rutherford In 1911, H.Geiger, and E Marsden performed an experiment called Geiger-Marsden experiment (or Rutherford scattering experiment)it consists of 1. Source of α particles : Poloniums is used as a source which placed in enclosure containing a hole and few slits 2. Thin gold foil: A gold foil nearly 10-8 m thick, α-particles are scattered by this foil.the whole experimental arrangement are so in the diagram. Observation and conclusions 1. Most of α particles pass through the gold foil undeflected.this shows that most part of the atom is hollow. 2. α particles are scattered through all angles.nearly 1 in 2000, suffer scattering though angles more than 90⁰,while smaller number (nearly I in 8000,) retrace path.this implies that when fast move positive charged α particles come near gold atom then a few of them experience such a strong repulsive force that they turn back. On this basis Rutherford concluded that whole of positive charge concentrated in small center called nucleus. The distance of closest approach estimated the nuclear size r 0 = 2Ze 2 /4π 0 E k Where symbols have their usual meaning Calculating so that size of nucleus is of the order of m,while size of atom is m.therefore size of nucleus is about /10-10 = 1/10,000 times the size of atom. 3. The negative charges (electrons) do not influence the scattering process, which show that whole mass of atom is concentrated in nucleus

5 Q.8. Using Bohr s postulates, of obtain an expression for the frequency of radiation emitted when electron make a transition from the higher energy state to lower energy state. Ans:- Let m = mass of an electron v = velocity of electron in n th orbit r = radius of the n th orbit The required centripetal force to the electron is provided by electrostatic force of attraction between electron and nucleus. mv 2 /r = 1/4πε 0 x (Ze) (e)/ r (i) Or, mv 2 = 1/4πε 0 x Ze 2 / r So, K. E. (K) = ½ mv 2 K = 1/4πε 0 x Ze 2 /2 r P. E. = 1/4πε 0 x (Ze) (-e)/ r = - 1/4πε 0 x Ze 2 / r Total Energy E = K. E. + P. E. = 1/4πε 0 x Ze 2 /2 r +( - 1/4πε 0 x Ze 2 / r) = - 1/4πε 0 x Ze 2 /2 r For n th orbit we can write E n = - 1/4πε 0 x Ze 2 /2 r n --- (ii) Now using Bohr s postulate mvr = nh/2π or, v = nh/2πmr Putting the value of v in equation (i) we get, m/r [nh/2πmr] 2 = 1/4πε 0 x Ze 2 / r 2 r = ε 0 h 2 n 2 /πmze 2 r n = ε 0 h 2 n 2 /πmze 2 Substituting value of r n in equation (ii), we get E n = - 1/4πε 0 x Ze 2 /2(ε 0 h 2 n 2 /πmze 2 ) = - mz 2 e 4 /8ε 0 h 2 n 2 E n = -Z 2 Rhc/n 2, where R = me 4 /8 ε 2 0 ch 3 (R = Rydberg constant) For hydrogen atom Z = 1 E n = -Rch/n 2 If n i and n f are the quantum numbers of initial and final states and E i and E f are energies of Electron in H- atom in initial and final state, we have 2 2 E i = -Rch/n i and E f = -Rch/n f If ν = frequency of emitted radiation, we get ν = E i E f /h = -Rc/n 2 i (-Rc/n 2 f ) = Rc[1/n 2 f 1/n 2 i ]

6 Q.9. State law of radioactivity and derive the relation N t = N 0 e - λ t for radioactive decay. Ans:- Radioactive Decay Law:- The rate of decay of radioactive nuclei is directly proportional to the number of undecayed nuclei at that time. Derivation of formula:- Let initially the number of atoms in radioactive element is N 0 and N the number of atoms after time t. After time t, let dn be the number of atoms which disintegrate in a short interval dt, then rate of disintegration will be dn/dt, this is also called the activity of the substance/element. According to Rutherford-Soddy law dn/dt N OrdN/dt = -λn --- (i) Whereλ is a constant of disintegration. Its unit is s -1. Negative sign shows that the rate of disintegration decreases with the increase of time. For a given element λ is constant. Equation (i) may be written as dn/n = -λdt Integrating, log N e = -λt + C --- (ii) Where C is constant of integration. At t = 0, N = N 0 Therefore, log e N 0 = 0+C => C = log e N 0 From equation (ii), log e N = -λt + log e N 0 or, log e N log e N 0 = -λt or, log e N/N 0 = -λt or, N/N 0 = e - λ t or, N = N 0 e -λt according to above equation, the number of undecayed atoms/nuclei of a given radioactive element decreases exponentially with time (i.e. more rapidly first and slowly afterwards).

7 Q.10. Define Half-life period and decay constant of a radioactive sample. Derive the relation between them. Ans:- Half-life Period (T) :- The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value. Decay Constant:- The decay constant of a radioactive element is defines as the reciprocal of time in which the number of undecayed nuclei of the radioactive element falls to 1/e times to its initial value. Relation:- We know the radioactive decay equation N = N 0 e -λt When t = T, N = N 0 /2 Therefore, N 0 /2 = N 0 e - λ T Or, e - λ T = ½ Taking log both sides -λt log e e = log e 1 log e 2 Or, λt = log e 2 Therefore, T = log e 2/λ = log 10 2/λ = x /λ Or, T = /λ Prepared By:- B. K. Singh PGT (Physics) K. V. Danapur Cantt (SS)

UNIT VIII ATOMS AND NUCLEI

UNIT VIII ATOMS AND NUCLEI Weightage Marks : 06 Alpha-particles scattering experiment, Rutherford s model of atom, Bohr Model, energy levels, Hydrogen spectrum. Composition and size of Nucleus, atomic