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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
In the carbon cycle form which hot starts obain their energy, the `._(6)^(12)C` nuclecus:A. is completely converted into energyB. is regenrated at the end of the cycleC. is broken up into its consituednt protons and neutronsD. is combined with oxygen to from carbon monoxide |
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Answer» Correct Answer - `(d)` It is fact |
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| 102. |
During a negative beta decay:A. An atomic electron is ejectedB. Am electron which is already present in nucleus is ejected.C. A neutron in the nucleus decay emitting electronsD. A part of the biniding energy of the nucleus is converted into electron |
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Answer» Correct Answer - `(c)` `._(0)^(1)n rarr ._(1)^(1)p + ._(-1)^(0)e + overline(v)` |
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| 103. |
The radioactive isotope `._(27)^(60)Co` which has now replaced radium in the treatment of cancer can be made by `a(n,p)` or `(n, gamma)` reaction. For each reactio, indicate the apporipriate target nucleus. If the half life of `._(27)^(60)Co` is 7 year evaluate the decay constant in `s^(-1)`. |
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Answer» `._(28)^(60)Ni + ._(0)^(1)n rarr ._(27)^(60) Co + ._(+1)^(1)P` `._(27)^(59) Co + ._(o)n^(1) rarr ._(27)^(60) Co + gamma` Thus, target nucleaus is `._(28)^(60)Ni` for `(n,p)` and `._(27)^(59)Co` for `(n, gamma)` reaction. `lambda = (0.693)/(t_(1//2))` `= (0.693)/(7xx365xx24xx60xx60)` `= 3.14xx10^(-9) sec^(-1)` |
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| 104. |
Caculalate the mass of `.^(140)LA` in a sample whose activity is `3.7xx10^(10)Bq` (`1` Becquerel, `Bq= 1` disntegration per second) given that is `t_(1//2)` is 40 hour. |
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Answer» Rate of decay `= lambda` `N_(0) = (lambda xx w xx Av. No .)/(140)` [Let `w` is weight of `La^(140)`, then `N_(0) = (w xx Av. No .)/(140)`] `3.7xx10^(10) = (0.693)/(40xx60xx60) xx (w xx 6.02xx10^(23))/(140)` `w = 1.79xx10^(-6) g` |
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| 105. |
The ratio of `C^(14)` to `C^(12)` in a living matter is measured to be `(C_(14))/(C_(12)) =1.3 xx 10^(-12)` at the present time Activity of `12.0gm` carabon sample is 180 dpm. The half life of `C^(14) ` is nearly _______ `xx 10^(12) sec`. [Given : `N_(A)=6 xx 10^(23)`]A. `0.18`B. `1.8`C. `0.384`D. 648 |
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Answer» Correct Answer - A Ratio of `C^(14) ` to `C^(12)` in living matter `=1.3 xx 10 ^(-12)` wt. of sample =12 g Total no. of atoms in sample `=(12)/(12) xx N_(A)=6.02 xx 10^(23)` (As the ratio of C-40 is quite lesser avg. atomic mass of sample can be taken as 12) No. of nuclei of C-14 in sample =No. of atoms of C-14 in sample `=6.02 xx 10^(23) xx 1.3 xx 10^(-12)` Activity = 180 dpm = 3 dps= `lambda` N `3=lambda xx 6.02 xx 10 ^(-23) xx 1.3 xx 10^(-12)` `lambda = (3)/(6.02 xx 10^(23) xx1.3 10^(-12))` `t_(1//2) = (0.693)/(lambda)=(0.693 xx 6.02 xx 10^(23) xx 1.3 xx 10 ^(-12))/(3) =0.18 xx 10^(12) sec`. |
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| 106. |
As an `alpha-` paricle approaches `a ._(7)N^(14)` nuclesus, the potential energy:A. Increases as it approaches nuclusB. Attains maximum value as inernuclear distance is approachedC. Reaches to a minium at the time of fusion.D. All of the above |
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Answer» Correct Answer - `(d)` Due to repulsion of `alpha-` particles with nucleous. |
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| 107. |
A solution of 1 litre has `0.6g` of non-radioactive `Fe^(3+)` with mass no. 56. To this solution `0.209g` of radioactove `Fe^(2+)` is added with mass no. 57 and the following reaction occurred. `.^(57)Fe^(2+) + .^(56)Fe^(3+) rarr .^(57)Fe^(3+) + .^(56)Fe^(2+)` At the end of one hour it was found that `10^(-5)` moles of non-radioactive `.^(56)Fe^(2+)` mol `L^(-1) hr^(-1)`. Negalecting any charge in volume, calculate the activity of the sample at the end of `1 hr` (`t_(1//2)` for `.^(57)Fe^(2+) = 4.62 hr`.) |
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Answer» `{:(,.^(57)Fe^(2+)+,.^(56)Fe^(3+)to,.^(57)Fe^(3+)+,.^(56)Fe^(2+)),("Before reaction",0.209/57,0.6/56,0,0),("After reaction",[0.209/57-10^(-5)],[0.6/56-10^(-5)],10^(-5),10^(-5)):}` `.^(57)Fe^(2+)` left after reaction `= 3.667xx10^(-3) - 10^(-5)` `= 366.6xx10^(5)` mole or `N_(0) = 366.6xx10^(-5)xx6.023xx10^(23) = 2.208xx10^(21)` Also `t = (2.303)/(lambda) log (N_(0))/(N)` `1 = (2.303xx4.62)/(0.693) log (2.208xx10^(21))/(N)` `N = 1.9xx10^(21)` `:.` Rate of decay `= lambda xx N = (0.693)/(4.62) xx 1.9xx10^(21)` `= 2.85xx10^(20)` dph |
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| 108. |
Most of the weight of the human body is because of the water present inside the body. Although various dissolved salts are also present, their concentration is very small and hence density of water inside the body can be taken as 1 gm /ml. A man weighing 100 kg is injected with a sample of radioactive water containing Tritium having activity `6.4 xx 10^(12) d` pm. After 10 min sample of water from the body is analysed and specific activity of water taken from body is found to be `(10^(6))/(27) dps//ml` Assuming the half life of the radioactive substance in water to be 2 min , calculate % by weight of water present in the human body. |
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Answer» Correct Answer - [0090] A man of weight 100 kg `A_(0)=6.4 xx 10^(12)d` pm `A_(t)=(6.4 xx10^(12))/(2^(5))d`pm `(6.4xx10^(12))/(2^(5)xx60)dps=(10^(6))/(27) dps//ml xx "Vml of water"` Vml=90000 `P_(H_(2)O)=1gm//ml` weight of `H_(2)O=90 kg` `% "of" H_(2)O` in body `=(90)/(100)xx100=0090` |
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| 109. |
The mean lives of radioactive substance are 1620 year and 405 year for `alpha-` emission and `beta-` emission respectively. Find out the time during which three fourth of a sample will decay, if it is decaying both by `alpha-` emission and `beta-` emission simulaneously. |
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Answer» Correct Answer - `449.24` year |
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| 110. |
Radioactive disintergation always follow `I` order kinetics and is independent of all external factors and is represented by the relation `N = N_(0) e^(-lambda t)` where `lambda` is decay constant and `N` atoms are left at time `t`. The radioactive nature of element is expressed in terms of average life numerically equal to decay constant `(1//lambda)` however all the radioactive do not lose their radioactive nature in thier average life. The radioactive emission involves `alpha, beta` particles as well as `gamma-` rays.The penertrating power order is `alpha lt beta lt gamma`. The emissions can perntrate even thick steel walls but are however unable to penttrate `Pb` blocks. The `S` unit fo rate of decay is `dps`. The percentage of atoms decayed in average life of a radioactive element is:A. `36.78%`B. `63.22%`C. `3.678%`D. `6.322%` |
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Answer» Correct Answer - `d` `t = (2.303)/(lambda) log (N_(0))/(N)`, `:. (1)/(lambda) = (2.303)/(lambda) log (N_(0))/(N), (t = t_(av.) = (1)/(lambda))` `:. (N_(0))/(N) = 2.71` or `N = 0.3678xx N_(0) = 36.78%` `:.` decayed `= 100 - 36.78 = 63.28` |
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| 111. |
In every 45 minutes, the number of nuclei of a radionuclides becomes 80 % of its number before 45 minutes, then the time (in minute) in which the number of nuclei of that nuclide becomes 64% of its initial number isA. 50B. 100C. 90D. 40 |
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Answer» Correct Answer - C Radioactive disintegration follow `I^(st)` order kinetics `lambda =(l)/(t) ln .(N_(0))/(N_(t)) " " (1)/(45) ln .(100)/(80) =(1)/(t) ln (100)/(64)` `t=45 ((2 ln 10 -6 ln 2)/(ln 10 -3 ln 2))=90` |
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| 112. |
Decreases in atomic number is observed in:A. `alpha-` emissionB. `beta-` emssionC. positron emssionD. electron capture |
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Answer» Correct Answer - `(a)` `alpha-` emission involves a decreases om atomic number by 2 units and mass number by 4 units. |
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| 113. |
A positron is emitted from `._(11)Na^(23)` . The ratio of the atomic mass and atomic number of the resulting nuclide isA. `22//10`B. `22//11`C. `23//10`D. `23//12` |
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Answer» Correct Answer - `(c)` During positon emssion, proton converts into neutron, therefore, atomic no, decreases by one but mass number remains constant. |
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| 114. |
Calculate the age of a vegetarian beverage whose tritium content is only 15% of the level in living plants. Given `t_(1//2) ` for `._(1) H^(3)= 12.3 ` years. |
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Answer» Correct Answer - `33.67 ` years `t=(1)/(lambda) ln. (%R.A.)/((%R.A.)"sample""` `t=(t_(1//2))/(0.693) ln. (x)/(x xx(15)/(100))=(12.3)/(0.693) ln. (20)/(3) rArr 33.67` year |
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| 115. |
The sun radiates energy at the rate of `4xx10^(26)` joule `sec^(-1)`. If the energy of fusion process `4 ._(1)^(1)H rarr ._(2)^(4)He + 2 ._(1)^(0)e` is `27MeV`, calculated amount of hydrogen atoms that would be consumed per day for the given process. |
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Answer» Correct Answer - `53.16xx10^(18)g` |
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| 116. |
In sun and other stars, where temperature is about `10^(7) K` fusion, takes place dominatly by:A. Proton-nitrogen cycleB. Proton-proton cycleC. proton-deuterium cycleD. proton-lithium cycle |
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Answer» Correct Answer - `(b)` It is a fact. |
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| 117. |
In botter star where the temperature is about `10^(8)K`, fusion takes place3 and the cyclic is known as:A. Proton-carbon cycleB. Proton-proton cycleC. Carbon-deuterium cylceD. Nitrogen-oxygen cycle |
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Answer» Correct Answer - `(a)` `-do-` |
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| 118. |
True or False Statements : The decay constant of the end product of a radioactive series is Zero |
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Answer» Correct Answer - [T] End product is last product i.e. , no furture decay |
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| 119. |
Statement -I : `._(6)^(14)C` is a `._(-1)^(0) beta` emitter. Statement -II Unstable nucleus having `n//p gt ` are `._(-1)^(" "0)beta` emitter.A. If both Statement -I & Statement -II are True & the Statement -II is a correct explanation of the Statement-I.B. If both Statement -I & Statement -II are True but Statement -II is not a correct explanation of the Statement-I.C. If Statement-I True but the Statement-II is FalseD. If Statement-I is False but the Statement-II is True. |
| Answer» Correct Answer - A | |
| 120. |
Statement-I : Coversionof a ` gamma` photon into an electron and a positron is an example of pair production. Statement -II : Pair production refers to the creation of an elementary particle and its antiparticle,usually when a photon interacts with a nucleus.A. If both Statement -I & Statement -II are True & the Statement -II is a correct explanation of the Statement-I.B. If both Statement -I & Statement -II are True but Statement -II is not a correct explanation of the Statement-I.C. If Statement-I True but the Statement-II is FalseD. If Statement-I is False but the Statement-II is True. |
| Answer» Correct Answer - A | |
| 121. |
If `E_(e)` is the energy needed to remove an electron from atom and `E_(n)` be energy needed remove a nucleion, then:A. `E_(n) lt E_(e)`B. `E_(n) gt E_(e)`C. `E_(n) lt E_(e)`D. `E_(n) ._(lt)^(gt) E_(e)` |
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Answer» Correct Answer - `(b)` Nucleions are more tightly held in nucleus due to strong nuclear forces. |
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| 122. |
Nuclear reacations either exoeerge or endoregic shows the exchange of:A. Kinetic energyB. Electrical energyC. Potential energyD. Heat energy |
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Answer» Correct Answer - `(a)` Nuclear reactions involving natural radioactivity (exoergic) or nuclear reactions following nuclear fission carried out by absorbing energy of slowly moving neutrons and then releasing energy only in form of kinetic energy. |
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| 123. |
Neutrions was predicted to:A. conserve mass of the nuclear reactionB. conserve charge of the nuclear reactionC. conserve spin of the nuclear reactionD. all of these |
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Answer» Correct Answer - `(c)` `._(0)^(0)v` proposed by Paulti to conserve spin in neutron decay. |
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| 124. |
Half-life-speed of lead is:A. infiniteB. `1590` dayC. `1590` yearD. zero |
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Answer» Correct Answer - `(a)` `Pb` is not radioactive. |
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| 125. |
`._(90)Th` a member of `III` group on losing `alpha-` particles forms a new element belonging to:A. `I` groupB. `III` groupC. `II` groupD. `IV` group |
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Answer» Correct Answer - `(c)` `._(90)^(232)Th rarr ._(88)^(228)Ra + ._(2)^(4)He`, Elements 89 to 103 are placed in `III` group. |
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| 126. |
`._(84)^(210)Po rarr _(82)^(206) Pb + ._(2)^(4)He` in this reaciton predict the positon of group of `Po` when lead is the the `IVB` group:A. `IIA`B. `IVB`C. `VIB`D. `VIA` |
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Answer» Correct Answer - `(c)` An element formed by losing one `alpha-` particles occupies two positon left to parent element. `Pb` in `IVB` thus `Po` thus `Po` sho db ein `ViB`. |
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| 127. |
Calculate the mass of oxalic `(H_(2)C_(2)O_(4).2H_(2)O)` which can be oxidised to `CO_(2)` by `100 mL` of `MnO_(4)^(-)` (acidic) solution, `10 mL` of which are capable of oxidising `50.0 mL` of `1.0 m I^(-)` to `I_(2)`. Also calculate the weight of `FeC_(2)O_(4)` oxidised by same amount of `MnO_(4)^(-)`. |
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Answer» Correct Answer - `H_(2)C_(2) O_(4) = 31.5g, FeC_(2)O_(4) = 24g` |
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| 128. |
Fill in the blanks with appropriate items : This binding energies per nucleon for `._(1)H^(2) ` and `._(2)H^(4)` are ` 1.1MeV and 7.0 MeV` respectively. The energy released when two deuterons fuse to form a helium nucleus is __________. |
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Answer» Correct Answer - `23.6 MeV` `2 ._(1)H^(2) rarr ._(2)He^(4)` `Delta E=rArr(2+2)(7)-2(1+1)(1.1)` `rArr 23.6 ` MeV |
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| 129. |
A `0.141g` sample of phosphorus containing compound was digested in a mixture of `HNO_(3)` and `H_(2)SO_(4)` which resulant in formnation of `CO_(2), H_(2)O` and `H_(3)PO_(4)`. Addition of ammounium molybdate yielded a solid having the composition `(NH_(4))_(3) PO_(4).12MoO_(3)`. The precipitate was filtered, washed and dissolved in `50.0mL` of `0.20M NaOH`. `(NH_(4))_(3) PIO_(4).12MoO_(3(s)) + 26OH^(-) rarr HPO_(4)^(2-) + 12MoO_(4)^(2-) + 14H_(2)O + 3NH_(3(g))` After boiling the solution to remove the `NH_(3)`, the excess of `NaOH` was titrated with `14.1 mL` of `0.174M, HCl`. Calculate the percentage of phosprous in the sample. |
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Answer» Correct Answer - `6.4%` |
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| 130. |
Loss of a `beta` -particle is equivalent toA. Increase of one proton onlyB. Decrease of one neutron onlyC. Both (A) and (B)D. None of these |
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Answer» Correct Answer - C `._(0)n^(1) rarr ._(1)P^(1)+._(-1)e^(0)` so loss of 1 `beta` particle is equivalent to decrease of one neutron or increase of one proton. |
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| 131. |
`_ (13)^(27) Al` is a stable isotope. `_(13)^(29) Al` is expected to disintegrated byA. `alpha` emissionB. `_(-1)^(0) beta` emissionC. Positron emissionD. Proton emission |
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Answer» Correct Answer - B Since `._(13)^(27) Al` is stable isotope , `._(13)^(29)Al` has `(n)/(P)` ratio higher than required, it will disintegrate by `beta^(-)` emission and `(n)/(P)` ratio will decrease. `._(13)^(29)Al rarr ._(14) si^(29)+._(-1)e^(0)` |
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| 132. |
`._(13)Al^(27)` is a stable isotope. `._(13)Al^(29)` is expected to disintegrate byA. `alpha-` emissionB. `beta-` emissionC. postion emissionD. proton emission |
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Answer» Correct Answer - `(b)` Nuclides having `(n)/(p) gt 1` undergoes `beta-` emission to decreases `(n)/(p)` in order to attain belt of stablitly i.e., `(n)/(p) = 1` `._(13)^(29)Al rarr ._(14)^(29)Si + ._(-1)e^(0)` |
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| 133. |
Which of the follwong nuclear reactiosn will generate an isotope?A. neutron particle emissionB. position emissionC. `alpha-` particle emssionD. `beta-` particle emission |
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Answer» Correct Answer - `(a)` `._(Z)^(A) X rarr ._(Z)^(A-1)X + ._(0)^(1)n` |
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| 134. |
The deacy time `t` for radioactive element proceeds to `4` half-lives. The total decay time `(t)` in terms of average life `(T)` is given by:A. `t = 27` In `2`B. `t = 4T` In `2`C. `t = 27^(4)` In `2`D. `(1)/(T^(2))` In `2` |
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Answer» Correct Answer - `(b)` `t = 4xx t_(1//2)` Also `t_(1//2) = (0.693)/(lambda) = 0.693xx T` `:. t = 4xx0.693xx T` `= 4T` in `2` |
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| 135. |
The half-life of `4.0mg beta-` emitter of `.^(210)X` is `5` day and the average energy of emitted `beta-` particle is `0.34MeV`. At what rate in watts does the sample emits energy?A. `2.0`B. `0.1`C. `1.5`D. `1.0` |
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Answer» Correct Answer - `(d)` Power = Energy of `beta` in `J xx "No of" beta-` particles emitted/sec. `= E xx (-(dN)/(dt)) = E xx lambda xx N` `= 0.34xx10^(6)xx1.6xx10^(-19) xx (0.693)/(5xx24xx60xx60)` `xx (4xx10^(-3)xx6.023xx10^(23))/(210) = 1J//sec = 1` watt |
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| 136. |
Which of the following statement is /are correct regarding nucleus of an atom ?A. As number of protons increases in the nucleus, the number of neutrons required for overcoming proton-proton repulsions also increase.B. Projectile capture reactions involving proton as projectile can occur with slow moving proton also.C. When `._(92)^(235) U ` undergoes fission reaction to from `._(36)Kr^(90) & ._(56)Ba^(142)` four neutrons are produced per Uraninum atom.D. Gamma radiations can be emitted only along with `alpha ` or `beta` - decay and not alone. |
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Answer» Correct Answer - A::C::D (A) Theory base (B) slow moving proton will not capture by the nucleus becose it will be repal be the nucleus (C) `._(92)^(235)U+._(0)n^(1) rarr .-(36)Kr^(90) +._(56)Ba^(142)+4_(0)n^(1)` (D) nucleus always release `gamma` ray during `alpha` , `beta` decay. |
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| 137. |
`1g Ra^(226)` is placed in an evacused tube whose volume is `5 cc`, Assuming that each `Ra` nucleus yields for `He` atoms which are retained in the tube, what will be the pressure of `He` preoduced at `27^(@)C` after the end of 1590 year? (`t_(1//2)` for `Ra` is `1590` year) |
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Answer» Correct Answer - `43.54` atm |
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| 138. |
For the following sequential reaction, `A overset(K_(1))(rarr)B overset(K_(2))(rarr)C` Find out the concentraion of `C` at tiem `t = 1` day. Given that `K_(1) = 1.8xx10^(-5) s^(-1)` and `K_(2) = 1.1xx10^(-2) s^(-1)` and initial molar concentration of `A` is `1.8` |
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Answer» Correct Answer - `1.42M` |
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| 139. |
Fill in the blanks with appropriate items : The number of ` alpha` and `beta ` - particles emitted, when the following nuclear transformation takes place are _______ and _______ respectively. `._(92)^(238) Xrarr ._(82)^(206)Y` |
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Answer» Correct Answer - 8,6 `._(92)^(238)X rarr. _(82)^(206)Y + x alpha + y alpha` `._(92)^(238)X rarr ._(82)^(206) Y + x(._(2)He^(4))+y(._(-1)^(0)e)` Comparing mass no. `238 rArr 206 + 4x +0y` `x=8 i.e. 8 (alpha) `particles Comparing no. protons. 92=82 + 2x-y `10=2 xx 8 -y` `y=6 i.e. 6 (beta)` particle |
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| 140. |
Fill in the blanks with appropriate items : When `._(c)^(a) X` changes to `._(d)^(b) Y` number if ` alpha` and `beta` particle emitted respectively are ______ |
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Answer» Correct Answer - `alpha=(a-b)/(4) ; beta = d+((a-b))/(2)-c` `._(c)^(a)M rarr ._(d)^(b)N+x(._(2)He^(4))+y(._(-1)e^(0))` `a=b+4x+0y` `(a-b)/(4)=x " " ..........(1)` `c=d+2x-y` `c-d-(a-b)/(2)=-y` `y=d+((a-b))/(2)=c` |
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| 141. |
Two radioactive elements `X` and `Y` half-live pf `50` and `100` minute respectiv ely. Intial sample of both the elements have same number of atoms. The ratio of the reamaining number of atoms of `X` and `Y` after 20 minute is:A. `2`B. `1//2`C. `4`D. `1//4` |
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Answer» Correct Answer - `(d)` For `X:200 = (2.303xx50)/(0.693) log ((a)/(x_(1)))` For `Y: 200 = (2.303xx100)/(0.693) log ((a)/(x_(2)))` `:. (x_(1))/(x_(2)) = (1)/(4)` |
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| 142. |
`._(92)^(235)U` belongs to `IIIB` group of the periodic table, It loses one `alpha-` particle, the new element will belong to the group.A. `IB`B. `IA`C. `IIB`D. `VB` |
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Answer» Correct Answer - `(c)` Elements 89 to 103 placed in `III` group. |
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| 143. |
A small subatojmic particle was passed through large watger tank conatining `Cd`. The existence of this particles was inferred when two`-gamma-` rays were produced and a neutron was captured by `Cd`. The particle was:A. protonB. neutrinosC. electronD. none of these |
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Answer» Correct Answer - `(b)` A fact about discovery of neutrions. |
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| 144. |
Which of the following statements is wrong?A. Area of cross-section of nucleas in about 1 barn `(1 "barn" = 10^(-24)cm^(2))`B. Elements placed below the belt of stablility shwo positorn emission to increase their `(n)/(p)` ratioC. Elements placed above the belt of stablility show `beta-` emission to decrease their `(n)/(p)` ratioD. `K-` electron capture emits `gamma` rays. |
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Answer» Correct Answer - `(d)` `K-` electron caputre always give `X-` rays due to jump of electron from higher energy level to `K-` shell in order to fill the vacancy created by `K-` electron capture. `._(37)^(81)Rb + ._(-1)^(0)e overset("K-electron caputure")(rarr) ._(26)^(81)Kr + "X-ray"` |
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| 145. |
Which of the following representations is correct ?A. `._(6)C^(12)(._(1)H^(1),. _(0)n^(1)) ._(7)N^(13)`B. `._(25)Mn^(55) (n,p) ._(25)Mn^(56)`C. `._(20)Ca^(40)(p,n) ._(21)Sc^(40)`D. `._(4)Be^(5)(p,n). _(3)Li^(6)` |
| Answer» Correct Answer - C | |
| 146. |
Which of the following statement is wrong?A. Nuclear radius is often expressed in fremi `(1F = 10^(-15)m)`B. Nuclear forces are short range and not strong attractive forcesC. Nuclear forces are about `10^(21)` times stronger than coulombic forcesD. Stabililty of nucleus is governed by inverse square law. |
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Answer» Correct Answer - `(d)` Stablilty of nuclieus is governed by nuclear forces and not by inverse square law i.e, coulombic forces of attractions. |
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| 147. |
Which of the following statements is wrong?A. `alpha-` decay always produces isodiapherB. `beta-` decay always produces isbarC. The maximum `(n)/(p)` ratio and maxiumum `(n)/(p)` ratio stands for `H-` isotoopesD. Synchroton can accelerate neutron particles. |
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Answer» Correct Answer - `(d)` Sychtron is accelerator charged particles only. |
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| 148. |
Some times a reacant undergoes chemical/radioactive changes following two or more different paths to yield two or more different produces respectively. Such reactions are called parallel path reactions. If `K_(1)` and `K_(2)` are rate constans for the reaction of `A` follwing two parallel paths, then Then `K_(av) = K_(1) + K_(2)` For if `E_(1)` and `E_(2)` are energy fo activations, thenA. `E_("Total") = E_(1) + E_(2)`B. `E_("Total") = E_(1) - E_(2)`C. `E_("Total") = K_(1) E_(1) + K_(2)E_(2)`D. `E_("Total") = (K_(1) E_(1) + K_(2)E_(2))/(K_(1) + K_(2))` |
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Answer» Correct Answer - `d` `K_(av) = K_(1) + K_(2)` `A_(T)e^(-E_(T)//RT^(2)) = A_(1) e^(-E_(1)//RT^(2)) + A_(2) e^(-E_(2)//RT^(2))` ....(1) Differentiating eqn. (1) `A_(T). (E_(T))/(RT).e^(-E_(T)//RT^(2)) = A_(1). (E_(1))/(RT).e^(-E_(1)//RT^(2)) + A_(2) (E_(2))/(RT) e^(-E_(2)//RT^(2))` `K_(av) (E_(T))/(RT) = (K_(1)E_(1))/(RT) + (K_(2) E_(2))/(RT)` `(K_(1) + K_(2)) (E_(T))/(RT) = (K_(1)E_(1))/(RT) + (K_(2) E_(2))/(RT)` `:. E_(T) = (K_(1) E_(1) + K_(2) E_(2))/(K_(1) + K_(2))` |
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| 149. |
A sample of `.^(14)CO_(2)` was to be mixed with ordinary `CO_(2)` for a biological tracer experiment . In order that `10 cm^(3)` of diluted gas should have `10^(4) "dis"//"min"`, what activity (in `muCi`) of radioactive carbon is needed to prepare 60L of diluted gas at STP. [`1Ci=3.7 xx10^(10)dps`] |
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Answer» Correct Answer - B A=`lambda`N,activity (or rate of decay)of 10 ml gas `rArr (10)^(4)/(60)`dps rate of decay of 60 litre gas`=((10^(4))/(60))xx(60xx1000)/(10)=10^(6)` sec `( :.1Ci=3.7xx10^(10dps)). " "=(10)^(6)/(3.7xx10^(10))=27xx10^(-6)Ci=27 mu Ci` |
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| 150. |
A certain radio isotope `._(Z)X^(A)` (half life `= 10 ` days) decays to give `._(Z-2)Y^(A-4)` . If `1.0 g` atom of `X` is kept in a sealed vessel, find the volume of helium accumulated at `STP` in `20` days ? |
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Answer» Correct Answer - 16.8L Decay of `._(Z)^(A)X " to " ._(Z-2)^(A-4)Y` produces one helium atom. Therefore, decay of 1.0g atom of `._(Z)^(A)X` will produce l.0g of atom of helium gas. Half-life of X=10days. Therefore, the amount decayed in w0 days`=(1)/(2)+(1)/(4)=(3)/(4)g ` of atom 1.0 g atom of He `=22400cm^(3) "at " 0^(@)C` and atm `3//4g` atom of He`=22,400xx(3)/(4)=16,800cm^(3)` |
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