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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.
| 51. |
Consider `alpha-, beta`- particles and `gamma-` rays, each having an energy fo `0.5Mev` in increasing order f penertation power, the radiations are:A. `alpha, beta, gamma`B. `alpha, gamma, beta`C. `beta, gamma, alpha`D. `gamma,beta,alpha` |
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Answer» Correct Answer - `(a)` `gamma` has more pentrating power |
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| 52. |
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 number of `beta-` particles emitted during the change, `._(a)^(c)X rarr ._(d)^(b) Y` is:A. `(a-b)/(4)`B. `d + [(a-b)/(2)] + c`C. `d + [(c - b)/(2)] -a`D. `d + [(a-b)/(2)] - c` |
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Answer» Correct Answer - `c` `._(a)^(c)X rarr ._(d)^(b)Y + n ._(2)^(4)Y + n ._(2)^(4)He + m._(-1) e^(0)` Equating mass no. `c = b + 4n + 0` `:. N = (c-b)/(4)` Equating at. No, `a = d + 2n + (-m)` `:. M = d - a + (2(c-b))/(4) = d - a + [(c-b)/(2)]` |
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| 53. |
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 completion of radioactive emission from a species takes place in:A. Average lifeB. Half-lifeC. `(1)/(2)xx`average lifeD. Infinity |
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Answer» Correct Answer - `d` First order reactions never goes for completion. |
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| 54. |
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`. Report the wrong realation:A. Amount decayed after `n` halves `= (N_(0) [2^(n) -1])/(2)`B. Av. Life `= t_(1//2) xx (1)/(0.693)`C. Fraction of nuclie decayed `= 1 - e^(-lambda t)`D. None of these |
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Answer» Correct Answer - `d` All are correct. |
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| 55. |
The nuclide ratio, `._(1)^(3) H` to `._(1)^(1) H` in a sample of water si `8.0xx10^(-18) : 1` Tritum undergoes decay with a half-line period of `12.3yr` How much tritum atoms would `10.0g` of such a sample conatins `40` year after the original sample is collected? |
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Answer» We know, `18g H_(2)O` has `2N H` atoms in it and `H^(3) : H^(1) : : 8xx10^(-18) : 1` `:. 18g H_(2)O` has `._(1)H^(3)` atoms `= 8xx10^(-18) x6.023xx10^(23) xx2` `:. 10g H_(2)O` has `._(1)H^(3)` atoms `= (8xx10^(-18)xx6.023xx10^(23)xx2xx10)/(18)` i.e., `N_(o) pf ._(1)H^(3) = 5.354 xx 10^(6)` atoms Now `t = (2.303)/(lamda) log_(10) (N_(o))/(N)` `40 = (2.303xx12.3)/(0.693) log_(10) (5.354xx10^(6))/(N)` `N = 5.624xx10^(5)` atoms |
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| 56. |
One mole of `A` present in a closed vessel undergoes decay as: `._(Z)^(A) rarr ._(Z-4)^(m-8)B + 2 ._(2)^(4)He`. The volume of `He` collected at `NTP` after 20 day (`t_(1//2)` for `A = 10` day) is:A. `11.2` litreB. `22.4` litreC. `33.6` litreD. `67.2` litre |
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Answer» Correct Answer - `(c)` Amount left `= (N_(0))/(2^(2)), (t_(1//2) xx n = T)` `:.` Amount decayed `= N_(0) - (N_(0))/(4) = (3N_(0))/(4)` Also 1 mole of `A` gives two moles of `He` thus moles of `He` formed `= (3)/(4)xx 2 = (3)/(2)` mole `= (3)/(2) xx22.4` litre `= 33.6` litre. |
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| 57. |
Equal masses of two samples of charcoal `A` and `B` are burnt separately and the resulting carbon dioxide is collected in two vessels. The radioacitivity of `.^(14)C` is measured for both the gas samples. The gas from the charcoal `A` gives `1400` counts per week. Find teh age difference between the two samples. (Half-life `.^(14)C = 5730yr`) |
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Answer» Correct Answer - `3353.16` year |
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| 58. |
When `._(17)Cl^(35)` undergoes `(n,p)` reaction, the radioisotope formed isa)`._(15)P^(32)`b)`._(16)S^(35)`c)`._(16)S^(34)`d)`._(15)P^(34)`A. `._(15)^(32)P`B. `._(16)^(35)S`C. `._(16)^(34)S`D. `._(15)^(34)P` |
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Answer» Correct Answer - `(b)` `._(17)^(35)C + ._(0)^(1)n - rarr ._(16)^(35)S + ._(1)^(1)H` |
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| 59. |
The half-life of a radioisotope is 1.5 hours. If its initial mass is 32g, the mass left undecayed after 6 hours isA. 32 gB. 16 gC. 4 gD. 2g |
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Answer» Correct Answer - D `N="mass left" =(N_(0)("initial mass"))/(2^(t)//^(t)_(1//2)` `N=32//2^(4)("since "t//_(t_(1//2))=6//_(1.5)=4)=2g` |
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| 60. |
The half-life of a radioactive isotope is 3 hour. IF the initial mass of isotope were `256g` the mass of it remaining undercayed after `18hr` is:A. `12g`B. `16g`C. `4g`D. `8g` |
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Answer» Correct Answer - `(c)` `N_("Left") = (N_(0))/(2^(n)) = (256)/(2^(6)) = 4g (T = t_(1//2) xx n)` |
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| 61. |
The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g , the undecayed mass remaining after 24 hours is :A. `1.042 g`B. `2.084 g`C. `3. 125 g`D. `4. 167 g` |
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Answer» Correct Answer - C `t_(1//2)=4` hours after 24 hours i.e. after 6 half times undecayed mass `=(200)/(2^(6))=(200)/(64)=3.125g` |
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| 62. |
The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the undecayed remaining mass after 18 hours would beA. `16.0g`B. `4.0 g`C. `8.0 g`D. `12. 0 g` |
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Answer» Correct Answer - B `t_(1//2)=3`hours after 18 hours,i.e. after 6 half lines undecayed remaining mass =`(256)/(2^(6))=(256)/(64)=4g` |
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| 63. |
Complete the following nuclear changes. (a) `._(42)^(96)Mo (..,n) ._(43)^(97)Tc` (b) `.. (alpha, 2n) ._(85)^(211)At` (c) `._(25)^(55)Mn (n, gamma)...` (d) `._(96)^(246)Cm + ._(6)^(12)C rarr ....+4 _(0)^(1)n` (e) `._(13)^(27) Al (alpha, n)..` (f) `._(235)^(92)U(alpha, beta^(-))..` |
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Answer» (a) `._(1)^(2)H`, (b) `._(83)^(209) Bi`, (c) `._(25)^(56)Mn`, (d) `._(102)^(254)No`, (e) `._(15)^(30)P`, (f) `._(95)^(242)Am` |
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| 64. |
Statement: The first man-made atom produced by artifical transmulation was `Tc`. Explanation: The phenolmenon of converting a stable nuclei into radioactive one is called artifical radioactivity.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - `d` Both facts are correct. |
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| 65. |
Predict the highest and lowest possible oxidation state of each of the following elements: (a) `Ta`, 9b) `Te`, (c) `Tc`, (d)`Ti`, (e) `Tl` (f) `N`, (g) `P`, `(h) `F`, (i) `Cl`, (j) `Zn` (k) `C`. |
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Answer» Correct Answer - (a) `5,0`, (b) `+6,2`, (c) `+7,0`, (d) `+4,0`, (e) `+3,0`, (f) `+5,-3`, (g) `+5, -3`, (h) `0,-1`, (i) `+7, -1`, (j) `+2,0`, (k) `+4,-4` |
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| 66. |
How may `alpha-` and `beta-` particles will be emitted when `._(90)Th^(232)` changes into `._(82)Pb^(208)`? |
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Answer» Suppose `._(90)Th^(232) rarr ._(82)Pb^(208) + m_(2)^(4) He + n_(-1)^(0)e` Equating mass numbers: `232 = 208 + 4m` or `m = 6` Equating atomic numbers: `90 = 82 + 12 + (-n)` or `n = 4` Thus, `6alpha` and `4beta^(-)` particles will be emitted. |
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| 67. |
Number of neutrons i8n a parent nucleus `X`, which gives `._(7)^(14)N` after two sucessive `beta-` emission would be:A. `6`B. `7`C. `8`D. `9` |
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Answer» Correct Answer - `(d)` `._(5)^(14)B rarr ._(7)^(14)N + 2 ._(-)^(0)e` |
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| 68. |
A radioactive element decays by `beta-` emission. If mass of parent and daughter nucliede are `m_(1)` and `m_(2)` respectively, caluclate energy liberated during the emission. |
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Answer» The masses of parent and daughter elements nuclide are `m_(1)` and `m_(2)` respectively. `._(2)^(X)A rarr _(Z + 1)^(X)B + ._(-1)^(0)e` Mass of parent element `= m_(1) + Zm_(e)` (where `Z` is at. No. of parent element.) Mass of daughter element `= m_(2) + (Z + 1)m_(e)` (As a. no fo element increases by one due to `beta-` emission) `:.` Mass decay = Mass of parent atom - One `beta` loss- [Mass of daugher atom along with one electron] `= m_(1) Zm_(e) - m_(e) - [m_(2) + (Z + 1) m_(e)]` `= [m_(1) - m_(2) - 2m_(e)]` Energy liberated `= [m_(1) - m_(2) - 2m_(e)] c^(2)` |
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| 69. |
Mass defect in the nuclear reactions may be expressed in terms of the atomic masses of the parent and daughter neucliders in place of their nuclear masses. The mass defect of nuclear reaction : `._(4)Be^(10)rarr ._(5)B^(10)+ e^(-)` isA. `Delta m="At. mass of " ._(4)Be^(10)-"At. mass of " _(5)B^(10)`B. `Delta m="At. mass of " ._(4)Be^(10)-"At. mass of " _(5)B^(10)` -mass of one electronC. `Delta m="At. mass of " ._(4)Be^(10)-"At. mass of " _(5)B^(10)` + mass of one electronD. `Delta m="At. mass of " ._(4)Be^(10)-"At. mass of " _(5)B^(10)` - mass of two electrons |
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Answer» Correct Answer - A Nuclear reaction `._(4)B^(10) rarr . _(5)B ^(10)+ e^(-)` Mass defect `=(4m_(p)+6m_(N))-[(5m_(p)+5m_(N))+m_(e)] ` `=(4m_(p)+6m_(N))-[(5m_(p)+5m_(N))+m_(e)] +4m_(e)-4m_(e)` `=(4m_(p)+6m_(N)+4m_(e))-[5m_(p)+5m_(N)+5m_(e)]` `="At mass of" ._(4)Be^(10)-"At. mass of " ._(5)B^(10)` |
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| 70. |
A radioactive sample emits n `beta`-particles in 2 sec. In next 2 sec it emits 0.75 n `beta` - particles, what is the mean life of the sample ? |
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Answer» Correct Answer - `1.75n=N_(0)(1-e^(-4 lambda)),6.95 sec, (2)/(ln((4)/(3)))` Let initial number of nuclei `=N_(0)` `N_(0)overset(2 "sec")rarr (N_(0)-n) overset(2 "sec")rarr(N_(0)-n-0.75 n)` (No. of `beta` particles emitted=No. of nuclei disintegrated) No. of nuclei disintegrated in time is given by `=N_(0)(1-e^(-lambdat t)` `:.` for initial 2 seconds `n=N_(0)(1-e^(-2lambda)) " " `...........(i) for next 2 seconds `0.75 n=(N_(0)-n)(1-e^(-2 lambda)) " " ` ..........(ii) Subtracting (ii) from (i) ((i)-(ii)) `.25 n=n (1-e^(-2lambda))` `e^(-2lambda)=0.75` `-2lambda=ln. (3)/(4) " " rArr 2lambda =ln .(4)/(3)` `t_("avg")=(1)/(lambda)=(2)/(ln((4)/(3)))=6.95 "sec"` |
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| 71. |
Fill in the blanks (a) `._(92)^(235)U+._(0)^(1)n rarr ._(52)^(137)A+ ._(40)^(97) B + __________. (b) `._(34)^(82) Se rarr 2 _(-1)^(0)e+__________`. |
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Answer» Correct Answer - (a) `2_(0)^(1)n` , (b) ` ._(36)^(82) Kr` `._(92)^(235)U+_(0)^(1)nrarr ._(52)^(137)A+_(40)^(97)B+X ._(0)n^(1)` (as the reaction is balanced with respect to nuclear charge, the missing particle must be neutral i.e. neutron) Applying mass nuclear charge balancing `34=-2+z` `z=36" "` (This implies element is kr) Applying mass number balancing `82=0+A` `A=82` Final balanced reaction is `._(34)^(82)Se rarr 2 ._(-1)^(0)e + _(36)^(82) Kr` |
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| 72. |
Corbon 14 is used to determine the age of organic material. The procedure is based on the formation of `.^(14)C` by neutron capture in the upper atmosphere. `._(7) N^(14) +._(0)n^(1) rarr. _(6)C^(14)+._(1)H^(1)` `.^(14)C` is absorbed by living organisms during photosynthesis. The `.^(14)C ` in the dead being falls due to the decay which `C^(14)` undergoes. `._(6) C^(14) rarr ._(7)N ^(14) + ._(-1)e^(@)` The half life period of `.^(14)C` is 5770 year. The decay constant `(lambda)` can be calculated by using the following formula `lambda= (0.693)/(t_(t//2))` The comparison of the `beta^(-)` activity of the dead matter with that of carbon still in circulation enables measurement of the period of hte isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of `.^(14)C to ._(12)C` in living matter is `1 : 10^(12)` What should be the age of fossil for meaningful determination of its age ?A. 6 yearsB. 6000 yearsC. 60000 yearsD. it can be used to calculate any age |
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Answer» Correct Answer - B As per the information given, the carbon dating method, ceases to be accurate over periods longer than `30,000 ` years. Hence the age of fossil for meaningful determination of its age should not be60000 years or 6 years (as it is quits lesser than `t_(1//2)`) The age of fossil should be near the `t_(1//2) ` of `c^(14)`.i.e. 6000 years. |
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| 73. |
Corbon 14 is used to determine the age of organic material. The procedure is based on the formation of `.^(14)C` by neutron capture in the upper atmosphere. `._(7) N^(14) +._(0)n^(1) rarr. _(6)C^(14)+._(1)H^(1)` `.^(14)C` is absorbed by living organisms during photosynthesis. The `.^(14)C ` in the dead being falls due to the decay which `C^(14)` undergoes. `._(6) C^(14) rarr. _(7)N ^(14) + ._(-1)e^(@)` The half life period of `^(14)C` is 5770 year. The decay constant `(lambda)` can be calculated by using the following formula `lambda= (0.693)/(t_(t//2))` The comparison of the `beta^(-)` activity of the dead matter with that of carbon still in circulation enables measurement of the period of hte isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of `.^(14)C to ._(12)C` in living matter is `1 : 10^(12)` Which of the following option is correct ?A. In living organisms, circulation of `.^(14)C` from atmosphere is high so the carbon content is constant in organismB. Carbon dating can be used to find out the age of earth crust and rocksC. Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organismD. Carbon dating cannot be used to determine concentration of `.^(14)C` in dead beings. |
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Answer» Correct Answer - C In living organism, `.^(14)C` content is constant (not carbon content) as the rate of absorption of `c^(14)` is equal to rate of disintegration of `.^(14)C`. Carbon dating can be used to fing out the age of organic materic only .(not for finding the age of earth crust and rocks made of inorganic material) |
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| 74. |
There is a stream of neutrons with KE=2eV. If the half life of neutrons is 100 sec, what fraction of neutrons will decay before they travel a distance of 1000 km. [Mass of neutrons `=1.6 xx10^(-24)g`]A. `0.75`B. `0.25`C. `0.707`D. `0.293` |
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Answer» Correct Answer - D `(1)/(2)MV^(2) rArr 2 eV` `V=sqrt((4eV)/(m))` `V=sqrt((4eV)/(1.6 xx 10^(-27))) " " l eV=1.6 xx 10^(-19)J` `sqrt((4 xx1.6 xx 10^(-19 )J)/(1.6 xx 10^(-27))) rArr V= sqrt(4 xx 10^(8))=2 xx 10^(4)m//s` `t=(1000 xx 1000m)/(V) rArr t= (10^(6)m)/(2 xx 10^(4))= 50 sec` `t=(t_(1//2))/(ln2) ln .(1)/((1-x)) rArr 50 = (100)/(ln2)ln .(1)/(1-x) rArr x=0.293` |
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| 75. |
How many atoms of `0.1 g`-atom of a radioacitve isotope `._(Z)X^(A)` (half life = 5 days) will decay during the 11th day? |
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Answer» `N_(0) = 0.1g` atoms, `t = 10` day and `t_(1//2) = 5` day `lambda = (2.303)/(t) log_(10) (N_(0))/(N)` `(0.6963)/(5) = (2.303)/(10) log_(10) (0.1)/(N)` `:. N_(10)` i.e., amount left after 10 day `0.0250g` atom Similarly, if `t=11day` `(0.693)/(5)=(2.303)/(11)log__(10)(0.1)/(N)` `:. N_(11)` i.e., amount left after 11 day `= 0.0218g` atom `:.` Amount decayed in `11th` day `= N_(10) - N_(11)` `= 0.0250-0.0218 = 3.2xx10^(-3)g` atoms `= 3.2xx6.023xx10^(23) xx 10^(-3)` atoms `= 1.93xx10^(21)` atoms |
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| 76. |
Helium nuclei combines to form an oxygen nucleus. The energy released per nucleon of oxygen nucleus is if `m_(0)=15.834` amu and `m_(He)=4.0026 `amuA. `10.24 MeV`B. 0MeVC. `5.24 MeV`D. 4 MeV |
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Answer» Correct Answer - A `4 ._(2)He^(4) rarr _(8)O^(16)` Mass defect `=4 xx 4.0026 -15.834 a.m.u.=0.1764 a.m.u.` Total energy released `=0.1764 xx 931 `MeV Energy released /Nucleon`=(0.1764 xx 931)/(16)=10.24 `MeV |
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| 77. |
The activity of a radioactive isotope falls to `12.5%` in 90 days. Compute the half life and decay constatn of isotope. |
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Answer» Correct Answer - `30 "day", 2.31xx10^(-2) "day"^(-1)` |
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| 78. |
The order of magnitude of density of urnaitum nucleus is: `(m_("nucleus") = 1.67xx10^(-27)kg)`A. `10^(20) kg//m^(3)`B. `10^(17) kg//m^(3)`C. `10^(14) kg//m^(3)`D. `10^(11) kg//m^(3)` |
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Answer» Correct Answer - `(b)` `m = A xx (1.67xx10^(-274))kg` (`A` is mass number) `V = (4)/(3) pi r^(3)` `= (4)/(3)pi [{1.3xx10^(-15)}A^(1//3)]^(3)` `= 8.2xx10^(-45) xx Am^(3)` `:. "Density = (m)/(V) = (1.67xx10^(-27)xx A)/(8.2xx10^(-45) xx A)` `= 2.0xx10^(17)kg//m^(3)` |
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| 79. |
The only stable isotope of sodium is `._(11)^(23)Na`. Daughter nucleus which can be formed due to radioactive decay of `._(11)Na^(22)` isA. `._(11)Na^(23)`B. `._(10)Ne^(22)`C. `._(9)F^(18)`D. `_(12)Mg^(22)` |
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Answer» Correct Answer - B `[._(11)Na^(23)` stable isotope i. e. `._(11)Na ^(22) ` has n/p ratio lesser than required To increase n/p ratio `{:(._(11)Na ^(22),rarr,._(10)Ne^(22),+,._(1)e^(0)),(,,,,("positron emission")):}` Or `{:(._(-1)e^(0),+,._(11)Na^(22),rarr,._(10)Ne^(22)),(("k-electron caputre"),,,,):}` |
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| 80. |
`._(11)^(23)Na` is the most stable isotope of Na. Find the process by which `._(11)^(24) Na` can undergo radioactive decay :A. `beta^(-)`-emissionB. `alpha`-emissionC. `beta^(+)`-emissionD. K-electron capture |
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Answer» Correct Answer - A Since `._(11)^(23)Na` is the stable isotope, in `._(11)Na^(24) ` n/p ratio is higher than required. It will decompose by `beta^(-)` emission , all the other three methods (`alpha` emission, `beta^(+)` emission, k-electron capture) increase n/p ratio. `{:(._(11)Na^(24),rarr,._(12)Mg^(24)+._(-1)e^(0),),(("unstable"),,("stable"),),(,,,):}` |
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| 81. |
Caluculate the density of uranium `-235` nucleus. Given `m_(n) = m_(p) = 1.67xx10^(-27) kg`. |
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Answer» Mass of uranium 235 = No. of nucleous `xxm_("nucleon")` `= A xx 1.67xx10^(-27)kg` Also volume of nucleus `= (4)/(3)pi r^(3)` `= (4)/(3) pi (1.3xx10^(-15)xx A^(1//3))^(3)` (where `A` is mass no.) `= 9.2xx10^(-45) xx A m^(3)` `:. "Density" = (m)/(V) = (1.67xx10^(-27)xx A)/(9.2xx10^(-45)xx A)` `= 1.8xx10^(17)kg m^(-3)` |
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| 82. |
Calucate the density of the nucelus of `._(47)^(107)Ag` assuming `R_("nucleus")` is `1.4A^(1//3)xx 10^(-13) cm`. Where `A` is mass number of nucelsus. Compare its density with density of metallic silver `10.5g cm^(-3)` |
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Answer» Volume of `._(47)Ag^(107)` nucleus `= (4)/(3)pi r^(3)` `= (4)/(3) xx (2)/(7) xx [1.4xx(107)^(1//3) xx 10^(-13)]^(3)` `= 1.23xx10^(-36) cm^(3)` `:.` Density of nucleus `= (m)/(V)` `= (107)/(6.023xx10^(23)xx1.23xx10^(-36))` `= 1.445xx10^(14)g//cm^(3)` Thus `("Density of nucleus")/("Density of atom") = (1.445xx10^(14))/(10.5)` `= 1.38xx10^(13)` |
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| 83. |
Write equations for the following transformations : (a)`._(7)N^(14)`(n,p), (b) `._(19)K^(39)(p,alpha)`, (c)`beta^(+)-"decay by" ._(11)Na^(22)` |
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Answer» (a) `._(7)N^(14)(n,p)` indicates that `N^(14)` on bombardment with neutrons gives proton. `:. " " ._(7)N^(14)+._(0)n^(1) rarr ._(z)X^(m)+._(1)p^(1)` on eqating at. "no". And mass "no". "on" both sides, we get `._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+ ._(1)p^(1)` (b) `._(19)K^(39)(p,alpha)` indicates that `K^(39)` on bombarment with proton gives `alpha `-particle. There, `._(19)K^(39)+._(1)H^(1) rarr ._(18)Ar^(36)+._(2)He^(4)` (c) `._(11)Na^(22) rarr . _(10)Ne^(22)+ ._(+1)e^(0)(beta^(+)"or positron")` |
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| 84. |
How much heat would be developed per hour form 1 curie of `C^(14)` source, if all the energy of beta decay were imprisoned? Atomic masses of `C^(14)` and `N^(14)` are `14.00324` absd `14.00307` amu respectively. |
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Answer» Correct Answer - `3.37J` |
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| 85. |
`._(6)^(11) C` on decay produces:A. PositronB. `beta-` particleC. `alpha-` particleD. `gamma-` rays |
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Answer» Correct Answer - `(a)` `._(6)^(11)C rarr ._(5)^(11) B + ._(+1)^(0)e` |
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| 86. |
One important source of energy of volacans erruption is:A. Hot mealn steam trapped in earthB. The pressure of ice at the earth poleC. Decay of radioactive matterD. The petroleum deposits stored under pressure |
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Answer» Correct Answer - `(c)` It is a fact. |
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| 87. |
In the nuclear chain ractio: `._(92)^(235)U + ._(0)^(1)n rarr ._(56)^(141) Ba + ._(36)^(92)Kr + 3 ._(0)^(1)n + E` The number fo neutrons and energy relaesed in nth step is:A. `3n, nE`B. `3^(n), nE`C. `3^(n), 3^(n-1) E`D. none of these |
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Answer» Correct Answer - `(c)` In `n^(th)` step neutrons ejected will be `3^(n)` and energy `3^(n-1)E` because one step ejects 3 electrons and energy `E.e.g` consider `II` step. In `2nd` step a neutrons `(3^(2))` and `3E` energy `(3^(2-1)E)` is liberated. |
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| 88. |
On igniion, Rochelle salt `NaKC_(4) H_(4) O_(6). 4H_(2)O` (mol. Wt 282) is converted into `NaKCO_(3) (mol wt. 122).0.9546g` sample of the rochelle salt on ignition gives `NaKCO_(3)` which is titrated wih `41.72mL.H_(2)SO_(4)`. From the follwing data, find the percentage purtiy of the rochelle salt. The solution after neutralisation requires its `1.91 mL` of `0.1297N NaOH`. The `H_(2)SO_(4)` used for the neutralisation requires its `10.27mL` aganist `10.35mL` of `0.1297N NaOH` |
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Answer» Correct Answer - `76.87%` |
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| 89. |
`0.108g` of finely divided copper was treated with an excess of ferric sulphate solution until copper was comnpletely dissolved. The solution after the addition of excess dilute sulphuric acid required `33.7mL` of `0.1N KMnO_(4)` for complete oxidation. Find the equaction which represents the reaction between metallic copper and ferric sulphate solution. At wt. of `Cu = 63.6` and `Fe = 56`. |
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Answer» Correct Answer - `Cu + Fe_(2)(SO_(4))_(3) rarr CuSO_(4) + 2FeSO_(4)` |
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| 90. |
In the nuclear transmutation : `._(4)^(9) Be + X rarr _(4) ^(8) Be +Y` (X,Y) is (are)A. `(gamma,n)`B. (p,D)C. (n,D)D. `(gamma,p)` |
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Answer» Correct Answer - A::B `._(4)^(9)Be rarr ._(4)^(8)Be` Check from option (A) `._(4)^(9) + lambda rarr ._(4)^(8)Be + ._(0)n^(1) " " (B) ._(4)^(9)Be + ._(2)^(1)Prarr ._(4)(8)Be+._(1)^(2)H` |
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| 91. |
The equation: `4._(1)^(1) H^(+) rarr ._(2)^(4) He^(2+) + 2e + 26MeV` represnets.A. `beta-` decayB. `gamma-` decayC. FusionD. Fission |
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Answer» Correct Answer - `(c)` It is a fact. |
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| 92. |
A sample of `U^(238)` (half-line `= 4.5xx10^(9) yr`) ore is found to contain `23.8g` of `U^(238)` and `20.6g` of `Pb^(206)` Calculate the age of the ore. |
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Answer» We know, `._(92)U^(238) rarr ``._(82)Pb^(206) + 8_(2) He^(4) + 6 ``._(-1)e^(0)` `Pb` present `= (20.6)/(206) = 0.1g` atom `= U` decayed `U` present `= (23.8)/(238) = 0.1g` atom, Thus, `N = 0.1g` atom `N_(0) = U` present `+ U` decayed `= 0.1+ 0.1 = 0.2g` atom Now `t = (2.303)/(k) log_(10) (N_(0))/(N)` `t = (2.303xx4.5xx10^(9))/(0.693) log_(10) (0.2)/(0.1)` `t = 4.5xx10^(9)` year |
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| 93. |
A sample of `U^(238) ("half life" = 4.5 xx 10^(9)yr)` ore is found to contain `23.8 g" of " U^(238) ` and `20.6g ` of `Pb^(206)`. Calculate the age of the ore |
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Answer» Correct Answer - [`4.5 xx 10^(9)` year] `U^(238) rarr Pb^(206)` `((20.6)/(206))=0.1 "mol present"` i.e. 0.1 mol of `U^(238)` decay `0.1 xx 238=(23.8)` decay Initial mass of `U^(238)=(23.8 +23.8)=47.6` 23.8 left and 23.8 decay time `t_(1//2)=4.5 xx 10^(9)` year |
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| 94. |
The analysis of a uranium reveals that ratio of mole of `.^(206)Pb`and `.^(238) U` in sample is 0.2 . If effective decay constant of process `.^(238)U rarr .^(206) Pb` is `lambda` then age of rock isA. `(1)/(lambda) In (5)/(4)`B. `(1)/(lambda) In ((5)/(1))`C. `(1)/(lambda) In(4)/(1)`D. `(1)/(lambda) In ((6)/(5))` |
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Answer» Correct Answer - D In the rock `("Moles of ".^(206)Pb)/("Moles of " .^(238 )U)=0.2=(0.2)/(1)` `{:(,.^(238)U,rarr,.^(206)Pb,),("Initial moles",1.2 "mole*",,0,"(*calculate by moles of"),(,1"mole",,0.2 "mole","U left & moles of Pb formed) "):}` `" " t=(1)/(lambda) ln ((1.2)/(1))=(1)/(lambda)ln ((6)/(5))` |
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| 95. |
A solution contains 1 milli-curie fo `L-` phenyol alanine `C^(14)` (unifromly labellled) in `2.0mL` solution. The activity of labelled sample is given as 150 milli-curie/milli-mole. Calculatate: (a) The concentration of sample in the solution in mol/litre. (b) The actitivity of the solution in terms of counting per minute/mL at a counting efficiency of `80%`. |
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Answer» Correct Answer - (a) `3.33xx10^(-2)M`, (b) `88.8xx10^(7) cpm//mL` |
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| 96. |
What mass of `C^(14)` with `t_(1//2) = 5730` years has activity equal to curie? |
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Answer» `-(dN)/(dt)=1 "curie"=3.7 xx 10^(10)` disintegrations per second `lambda=(0.693)/(5730xx365 xx24 xx3600)` `-(dN)/(dt)=lambda N " or " 3.7 xx10^(10)=(0.693)/(5730 xx365xx24xx3600)xxN` Or `N=9.65xx10^(21)` atoms Or amount of `.^(14)C=(14)/(6.023 xx10^(23)) xx9.65 xx10^(21)=0.2244g` |
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| 97. |
A solution contains a mixture of isotopes of `X^(A)`(`t_(1//2) = 14` days) and `X^(A_(2))` (`t_(1//2) = 25` days(. Total acticity is 1 curie at `t = 0`. The activity reduces by `50%` in 20 days. Find: (a) The initial activitites of `X^(A_(1))` and `X^(A_(2))` (b) The ratio of their initial no.of nuclei. |
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Answer» Correct Answer - (a) `3.3669 Ci, 0.6331 Ci`, (b) `0.3245` |
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| 98. |
The number of millimoles of `._(6)^(14)C` equivalent to one millicuires if `t_(1//2) = 5770` year and `1` curie `= 3.7xx10^(10)` dps is:A. `1.56xx10^(-2)`B. `3.12xx10^(-2)`C. `4.34xx10^(-2)`D. `7.80xx10^(-2)` |
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Answer» Correct Answer - `(a)` 1 milli curie `= 3.7xx10^(10)xx10^(-3)` dps `:. r = lambda.N` or `3.7xx10^(7) = (0.693)/(5570xx365xx24xx60xx60) xx N` `:. N = 9.38xx10^(18)` `:.` Mole of `._(6)^(14)C = 1.56xx10^(-2)` |
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| 99. |
The radioactive nuclide `._(90)^(234) Th` shows two successive `beta-` decay followed by one `alpha-` decay. The atomic number and mass number respectively of the resulting atom is:A. `90` and `230`B. `92` and `230`C. `92` and `234`D. `94` and `230` |
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Answer» Correct Answer - `(a)` `._(90)^(234) Th rarr ._(90)^(230)Th + ._(2)He^(4) + 2 ._(-1)e^(0)` |
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| 100. |
Wooden article and freshly cut tree show activity of `7.6` and `15.2 min^(-1) gm^(-1)` of carbon (`t_(1//2)=5760` years ) respectively. The age of article in years, isA. 5760B. `5760 xx ((15.2)/(7.6))`C. `5760 xx ((7.6)/(15.2))`D. `5760 xx (15.2-7.6)` |
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Answer» Correct Answer - A Since the activity of wooden article is half of activity of freshly cut free, age of the article is `t_(1//2)=5760` years |
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