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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. |
The `n//p` ratio for a stable lighter nuclei is about `………………` . |
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Answer» Correct Answer - One |
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| 102. |
One microgram of phosphorus-32 was injected into a live syetm for biological studies. The half life `""_(15)P""^(32)` is 14.3 days, calculate the time it will take the radioactivity to fall to 10% of the initial valueA. 47.52 daysB. 57.52 daysC. 100 daysD. 4.7 days |
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Answer» Correct Answer - a `lamda=2.303/tlog.N""_(0)/N` |
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| 103. |
In the reaction `._1^2 H + ._1^3 H rarr ._2^4 He + ._0^1 n`, if the binding energies of `._1^2 H, ._1^3 H` and `._2^4 He` are respectively `a,b` and `c` (in MeV), then the energy (in MeV) released in this reaction is.A. `a +b -c`B. `c + a -b`C. `c -a -b`D. `a + b + c` |
| Answer» Correct Answer - C | |
| 104. |
Half life of a radioactive sample is `2x` years. What fraction of this sample will remain undecayed after `x` years? |
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Answer» `K = (2.303)/(t) log ((N_(0))/(N))` `(0.693)/(t_(1//2)) = (2.303)/(t) log ((N_(0))/(N))` `(0.693)/(2x) = (2.303)/(x) log ((N_(0))/(N))` `(1)/(2) log 2 = log ((N_(0))/(N))` `(N)/(N_(0)) = (1)/(sqrt(2))` Fraction undecayed `= (1)/(sqrt(2))` |
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| 105. |
80% of the radioactive nuclei present in a sample is found to remain undecayed after one day. The percentage of undecayed nuclei left after two days will be a. 64 b. 20 c. 46 d. 80A. a. 64B. b. 20C. c. 46D. d. 80 |
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Answer» Correct Answer - a. 64 `lambda = (2.303)/(t) log ((N_(0))/(N))` `= (2.303)/(1) log ((100)/(80)) [t = 1 "day"]` `lambda = (2.303)/(2) log ((100)/(80)) [t = 2 "days"]` `(2.303)/(1) log ((100)/(80)) = (2.303)/(2) log ((100)/(N))` `((5)/(4))^(2) = (100)/(N)` `N = 64` |
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| 106. |
One mole of a present in a closed vessel undergoes decay as `""_(z)""^(m)Ato""_(Z-4)""^(m-8)B+2""_(2)""^(4)He`. The volume of He collected at NTP a fter 20 days is (`t""_(1//2)=10`days)A. 11.2 litreB. 22.4 litreC. 33.6 litreD. 67.2 litre |
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Answer» Correct Answer - c Amount of (A)left=`N""_(0)/2""^(2)(t""_(1//2)xxn=T)` Amount of (A) decayed=`N""_(0)-N""_(0)/2=(3N(0))/4` `therefore` Mole of He formed=`2xx3/4N""_(0)` `=3/2N""_(0)=3/2`mole(`because` 1 mole of (A) is taken) `therefore` volume of He =`3/2xx22.40=33.6` litre |
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| 107. |
Assertion : Nuclear forces and called short range forces. Reason :Nuclear forces operate over very small distance i.e., `10^(-15)` m or 1 fermiA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanantion of the assertionC. If assertion is true but reason is falseD. If the assertion and reason both are false |
| Answer» Correct Answer - A | |
| 108. |
What kind of radioactive decay does not lead to the formation of a daughter nucleus that is an isobar of the parent nucleusA. `alpha-`raysB. `beta-`raysC. PositronD. Electron capture |
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Answer» Correct Answer - A When a radioactive element emits an `alpha`-particle, the atomic no. of the resulting nuclide decreases by two units and atomic mass decreases by 4 units |
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| 109. |
The radioaisotope of hydrogen has a half-life of 12.33 y. What is the age of an old bottle of wine whose `._(1)^(3)H` radiation is 10% of that present in a new bottle of wineA. 41yB. 12.3 yC. 1.233 yD. 410 y |
| Answer» Correct Answer - A | |
| 110. |
Given that a radioactive species decays according to exponential law `N= N_(0) e^(-lamda t)`. The half-life of the species isA. `lamda`B. NoC. `lamda//ln 2`D. `ln 2//lamda` |
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Answer» Correct Answer - D `t_(1//2) = ln 2//lamda` |
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| 111. |
Shorter the radioactive half lifeA. Greater is the number of atoms disintegrating per secondB. Smaller is the decay constantC. Less reactive is the parent nucleusD. Greater is the mass -energy change |
| Answer» Correct Answer - A | |
| 112. |
If `N_(0) and N` are the number of radioactive particles at time `t = 0 and t = t`, thenA. `lamda = (1)/(t) "log" (N_(0))/(N)`B. `lamda = (2.303)/(t) "log"(N)/(N_(0))`C. `lamda = (t)/(2.303)"log"(N_(0))/(N)`D. `lamda = (2.303)/(t) "log" (N_(0))/(N)` |
| Answer» Correct Answer - D | |
| 113. |
Two radioactive elements X and Y have half-lives of 6 min and 15 min respectively. An experiment starts with 8 time sas many atoms of X as Y. How long it takes for the number of atoms of X left equals the number of atoms of Y leftA. 6 minB. 12 minC. 48 minD. 30 min |
| Answer» Correct Answer - D | |
| 114. |
The half life of radioactive isotope is 3 hour. If the initial mass of isotope were 256 g, the mass of it remaining undecayed after 18 hr isA. 12 gB. 16 gC. 4 gD. 8 g |
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Answer» Correct Answer - c `N""_("left")=(N""_(0))/(2""_(n))=256/2""^(6)=4g(T=t""_(1//2)xxn)` |
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| 115. |
A substance is kept for 2 hours and three-fourth of that substance disintegrates during this period. The half life of the substance isA. 2 hrB. 1 hrC. 30 minD. 4 hr |
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Answer» Correct Answer - B Given, Amount left `N = 1 - (3)/(4) = (1)/(4)` `N = N_(0) ((1)/(2))^(n) implies ((N)/(N_(0))) = ((1)/(2))^(n)` `implies ((1)/(4)) = ((1)/(2))^(n)` or `n = 2` we know `T = n xx t_(1//2) implies t_(1//2) = (2)/(2) = 1 hr` |
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| 116. |
What is the half-life of a radioactive substance if 75% of a given amount of the substance disintegrates in 30 minutesA. 7.5 minutesB. 25 minutesC. 20 minutesD. 15 minutes |
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Answer» Correct Answer - D 75% of the substance disintegrates in two half lives 2 half lives = 30 min `:. t_(1//2)` = 15 min |
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| 117. |
A sample of radioactive substance shows an intensity of 2.3 millicurie at a time t and an intensity of 1.62 millicurie after 600 s. The half-life period of the radioactive metal isA. 1000 sB. 1187 sC. 1200 sD. 1500 s |
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Answer» Correct Answer - b `(N_(0))/(N)=(2.30)/(1.62)`, Now, `lambda = (2.303)/(600)log(2.30)/(1.62) = 0.000584` `:. T_(1//2) = (0.693)/(lambda) = (0.693)/(0.000584) = 1187s` |
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| 118. |
Assertion `(A):` Half life of a radioactive isotope is the time required to decrease its mass number by half. Reason `(R) : ` Halt`-` of readioactive isotopes is independent of the initial amount of the isotope.A. If both `(A)` and `(R)` are correct , and `(R)` is the correct explanation of `(A)`B. If both `(A)` and `(R)` are correct, but (R) is not the correct explanation of `(A)`C. If `(A)` is correct, but `(R)` is incorrect.D. If both `(A)` and `(R)` are incorrect. |
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Answer» Correct Answer - D Half`-` life is the time period required to decrease the initial number of atoms or amount of radioactive substance to half. |
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| 119. |
The half`-`life of a radioactive isotope is always. `…………………………`. |
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Answer» Correct Answer - Constant |
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| 120. |
If `3//4` quantity of a radioactive substance disintegrates in 2 hours, its half`-` life period will beA. `15 mi n`B. `30 mi n`C. `60 mi n`D. `90 mi n` |
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Answer» Correct Answer - C `t_(3//4)=2t_(1//2)` `:. t_(1//2)=(t_(3//4))/(2)=(2xx60)/(2)=60 mi n` |
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| 121. |
Assertion (A) : Half life of a radioactive isotope is the time required to decrease its mass number by half Reason (R ) : Half life of radioactive isotope is independent of the initial amount of the isotopeA. If both (A) and (R ) are correct, and (R ) is the correct explaination of (A)B. If both (A) and (R ) are correct, but (R ) is not the correct explaination of (A)C. If (A) is correct,but (R ) is incorrectD. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - c | |
| 122. |
The wavelength of the radiation emitted , when in a hydrogen atom electron falls from infinity to stationary state 1 , would be : (Rydberg constant = `1.097 xx 10^(7) m^(-1)`)A. `91 nm`B. `192 nm`C. `406`D. `9.1xx10^(-6) nm` |
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Answer» Correct Answer - 1 `1/lambda=R(1/n_(1)^(2)-1/n_(2)^(2)) 1/lambda=1.097xx10^(7) m^(-1)(1/1^(2)-1/oo^(2)) :. lambda=91xx10^(-9) m=91 m` |
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| 123. |
Assertion `:(A): ` The average life of radioactive element is infinity. Reason `(R):` As a radioactive element disintegrates, more of it is formed in nature by itself.A. If both `(A)` and `(R)` are correct , and `(R)` is the correct explanation of `(A)`B. If both `(A)` and `(R)` are correct, but (R) is not the correct explanation of `(A)`C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, but `(R)` is correct. |
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Answer» Correct Answer - C The complete disintegration of any radioactive material is not possible. |
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| 124. |
Assertion (A) : The average life of a radioactive element is infinity Reason (R ) : As a radioactive element disinegrates, more of it is formed in nature by itselfA. If both (A) and (R ) are correct, and (R ) is the correct explaination of (A)B. If both (A) and (R ) are correct, but (R ) is not the correct explaination of (A)C. If (A) is correct,but (R ) is incorrectD. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - c | |
| 125. |
The phenomenon of radioactivity is associated withA. Decay of nucleusB. Fussion of nucleusC. Emission of electrons or protonsD. Rearragement in the in the extra nuclear electron |
| Answer» Correct Answer - A | |
| 126. |
Which of the following statements about radioactivity is are true?A. It involves outer electrons activity.B. It is not affected by temperature of pressure.C. It is an exothermic process.D. The radioactivity of an element is not affected by any other element compounded by it. |
| Answer» Correct Answer - B::C::D | |
| 127. |
Radioactivity is due toA. Stable electronic configurationB. Unstable electronic configurationC. Stable nucleusD. Unstable nucleus |
| Answer» Correct Answer - D | |
| 128. |
Alpha rays areA. Positively chargedB. Negatively chargedC. NeutralD. |
| Answer» Correct Answer - A | |
| 129. |
Given that the abundances of isotopes `.^(54)Fe, .^(56)Fe and .^(57)Fe` are 5%, 90% and 5% respectively, the atomic mass of Fe isA. 55.85B. 55.95C. 55.75D. 56.05 |
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Answer» Correct Answer - B `.^(54)Fe rarr 5%` `.^(56)Fe rarr 90%` `.^(57)Fe rarr 5%` Av. Atomic mass `= x_(1) A_(1) + x_(2) A_(2) + x_(3) A_(3)` `= 54 xx 0.05 + 56 xx 0.9 + 57 xx 0.05 = 55.95` |
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| 130. |
A solution contains `1 mCi` of `L`-phenylalanine `C^(14)` labelled in `2.0 mL` solution. The specific activity of labelled sample is given as `150 mCi "mmol"^(-1)`. Calculate (a) The concentration of the sample in the solution in `"mol" L^(-1)` (b). The activity of solution in terms of counting per minute per `mL` at counting of 80% |
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Answer» a. 1 mmol ` = 150 mCi` `1 mCi = (1)/(150)` mmol `= (1)/(150 xx 2) = 3.33 xx 10^(-2) M` b. `1 Ci = 3.7 xx 10^(10) dps` `= 3.7 xx 10^(10) xx 60 dpm` `= 3.7 xx 10^(10) xx 60 xx (80)/(100)` counts `"min"^(-1)` `= (177.6 xx 10^(10))/(2 mL) = 88.8 xx 10^(10)` counts `"min"^(-1) mL^(-1)` ` mCi = 88.8 xx 10^(7)` count `"min"^(-1) mL^(-1)` |
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| 131. |
An element `X` loses one `alpha-` and two `beta-` particles in three successive stages. The resulting element will beA. An isobar of `X`B. An isotope of `X`C. `X` itselfD. An isotone of `X` |
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Answer» Correct Answer - B Let the reaction is `._(Z)X^(A) overset(-alpha)rarr ._(Z-2)A^(A-4)overset(-beta)rarr ._(Z-1)B^(a-4) overset(-beta)rarr ._(Z)C^(A-4)` Therefore, `X` and `C` are isotope `i.e.,` same atomic number. |
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| 132. |
`a. ._(92)U^(235)+._(0)n^(1)rarr (52)A^(137)+._(40)B^(97)+..........` `b. ._(34)Se^(84)rarr 2 ._(-1)e^(0)+..........`A. Statement I is true , Statement II is true, Statement II is the correct explanation of Statement IB. Statement I is true, Statement II is true, Statement II is not the correct explanation of Statement IC. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true |
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Answer» Correct Answer - `2 _(0)n^(1),_(36)Kr^(82)` |
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| 133. |
Decrease in atomic number is observed duringA. `alpha`-emissionB. `beta`-emissionC. positron emissionD. electron capture |
| Answer» Correct Answer - a,c | |
| 134. |
Decrease in atomic number is not observed inA. `alpha`-emissionB. `beta`-emissionC. positron emissionD. electron emission |
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Answer» Correct Answer - b Atomic No increased by one unit by `beta` emission |
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| 135. |
Decrease in atomic number is observed during :A. Alpha emissionB. Beta emissionC. Positron emissionD. Electron emission |
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Answer» Correct Answer - A::C::D `alpha` emission : `._(Z)X^(A)rarr ._(Z-2)Y^(A-4)+ ._(2)He^(4)` Positron emission : `._(Z)X^(A)rarr ._(Z-1)Y^(A)+ ._(+)e^(0)+V` Electron capture : ` ._(1)p^(1)+ ._(-1)e^(0) rarr ._(0)n^(1)+V` |
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| 136. |
An isotone of `._(32)^(76)Ge` is (one or more are correct)A. `._(32)^(77)Ge`B. `._(33)^(77)As`C. `._(34)^(77)Se`D. `._(34)^(78)Se` |
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Answer» Correct Answer - B::D Both have 44 neutrons, Isotones have same number of neutrons. |
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| 137. |
Assertion (A) : Nuclide `AI_(13)^(30)` is less stable than `Ca_(20)^(40)` Reason (R ) : Nuclide having odd number of proton and neuctrons are generally unstableA. Statement I is true , Statement II is true, Statement II is the correct explanation of Statement IB. Statement I is true, Statement II is true, Statement II is not the correct explanation of Statement IC. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true |
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Answer» Correct Answer - B |
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| 138. |
Decrease in atomic number is observed during :A. alpha emissionB. beta emissionC. positron emissionD. electron capture |
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Answer» Correct Answer - `(A,B,C)` |
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| 139. |
Decrease in atomic number is observed duringA. Alpha emissionB. Beta emissionC. Positron emissionD. Electron capture |
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Answer» Correct Answer - A::C::D Beta emission causes increases in atomic number by one unit |
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| 140. |
The nuclear reactions accompanied with emission of neutron `(s)` areA. `._(13)^(27)Al+_(2)^(4)Heto_(15)^(30)P`B. `._(6)^(12)C+_(1)^(1)Hto_(7)^(13)N`C. `._(15)^(30)Pto_(14)^(30)Si+_(1)^(0)e`D. `._(96)^(241)Am+_(2)^(4)Heto_(97)^(244)Bk+_(1)^(0)e` |
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Answer» Correct Answer - `(A,D)` |
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| 141. |
Nuclear reactions accompanied with emission of neutron (s) areA. `._(13)^(27)Al + ._(2)He^(4) rarr ._(15)^(30)P`B. `._(6)^(12)C + ._(1)^(1)H rarr ._(7)^(13)N`C. `._(15)^(30)P rarr ._(14)^(30)Si + ._(1)^(0)e`D. `._(96)^(241)Am + ._(2)^(4) He rarr ._(97)^(244)Bk + ._(1)^(0)e` |
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Answer» Correct Answer - A::D (a) `._(13)Al^(27) + ._(2)He^(4) rarr._(15)P^(30) + ._(0)n^(1)`, (b) `._(96)Am^(241) + ._(2)He^(4) rarr ._(97)BK^(244) + ._(1)e^(0) + ._(0)n^(1)` |
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| 142. |
Positron emission results from the transformation of one nuclear proton into neutron. The isotope thus produced possessesA. Same mass numberB. Higher nuclear chargeC. Intense radioactivityD. No radioactivity |
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Answer» Correct Answer - A Mass no. will remain same as proton is replaced by neutron |
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| 143. |
Carbon`-14` used to determine the age of organic material. The procedure is absed on the formation of `C^(14)` by neutron capture iin the upper atmosphere. `._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1)` `C^(14)` is absorbed by living organisms during photosynthesis. The `C^(14)` content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of `C^(14)` in the dead being falls due to the decay, which `C^(14)` undergoes. `._(6)C^(14)rarr ._(7)N^(14)+beta^(c-)` The half`-` life period of `C^(14)` is 5770 year. The decay constant `(lambda)` can be calculated by using the following formuls `:` `lambda=(0.693)/(t_(1//2))` The comparison of the `beta^(c-)` activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of `C^(14)` to `C^(12)` 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. 60000yearsD. It can be used to calculate any age |
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Answer» Correct Answer - B Radiacarbon dating method ceases to be accurate over periods longer than 30000 years. |
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| 144. |
Carbon`-14` used to determine the age of organic material. The procedure is absed on the formation of `C^(14)` by neutron capture iin the upper atmosphere. `._(7)N^(14)+ ._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1)` `C^(14)` is absorbed by living organisms during photosynthesis. The `C^(14)` content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of `C^(14)` in the dead being falls due to the decay, which `C^(14)` undergoes. `._(6)C^(14)rarr ._(7)N^(14)+beta^(c-)` The half`-` life period of `C^(14)` is 5770 year. The decay constant `(lambda)` can be calculated by using the following formuls `:` `lambda=(0.693)/(t_(1//2))` The comparison of the `beta^(c-)` activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of `C^(14)` to `C^(12)` in living matter is `1:10^(12)`. Which of the following options is correct ?A. In living organisms, circulation of `.^(14)C` from the atmosphere is high, so the carbon content is constant in organism.B. Carbond dating can be used to find out the age of earth crust and rocks.C. Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay. Hence, the carbon content remains constant in living organisms.D. Carbon dating cannot be used to determine concentration of `C^(14)` in dead beings. |
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Answer» Correct Answer - C Radioactive absorption due to cosmic radiation is equal to the rate of radiactive decay, hence the carbon content as the ratio of `C^(14)` remains constant in living organism. |
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| 145. |
The nuclear reaction `(s)` accompanied with the emission of neutron`(s)` is `//` areA. `._(13)Al^(17)+._(2)He^(4)rarr ._(15)P^(30)`B. `._(6)C^(12)+._(1)He^(1) rarr._(7)N^(13)`C. `._(15)P^(30)rarr ._(14)Si^(30)+._(1)e^(0)`D. `._(96)Am^(241)+._(2)He^(4)rarr ._(97)Bk^(244)+._(1)e^(0)` |
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Answer» Correct Answer - A::D `._(13)Al^(27)+._(2)He^(4) rarr ._(15)P^(30)+._(0)n^(1)` `._(96)Am^(241)+._(2)He^(4)rarr ._(97)Bk^(244)+._(1)e^(0)+._(0)n^(1)` |
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| 146. |
The emission of an `alpha or a beta` particle by a radioactive element forms a new element. However, successive emission of some `alpha or beta`- particles may give rise to an isotope or an isobar of the original element. In many cases, positron emission or `K -` electron capture takes place, leading again to the formuation of new elements, alongwith the emission of neutrinos or antineutrinos. These emission also change the neutron/proton `(n//p)` ratio so that they give rise to stable isotopes which lie in the stability belt. However, in any disintegration reaction, the law of conservation of atomic number and mass number is always obeyed and this helps us to calculate the number of `alpha and beta-` particles emitted in the reaction. The number of `alpha - and beta-` particle emitted in nuclear reaction `._(90)Th^(288) rarr ._(83)Bi^(212)` are respectivelyA. 4, 1B. 3, 7C. 8, 1D. 4, 7 |
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Answer» Correct Answer - A `._(9)Th^(228) rarr ._(83)Bi^(212) + x_(2)^(4) alpha + y ._(-1)beta^(0)` `228 = 212 + 4x or x = 4` `90 = 83 + 2x - y or y = 1` |
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| 147. |
Carbon`-14` used to determine the age of organic material. The procedure is absed on the formation of `C^(14)` by neutron capture iin the upper atmosphere. `._(7)N^(14)+._(0)n^(1) rarr ._(6)C^(14)+._(1)H^(1)` `C^(14)` is absorbed by living organisms during photosynthesis. The `C^(14)` content is constant in living organism. Once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of `C^(14)` in the dead being falls due to the decay, which `C^(14)` undergoes. `._(6)C^(14)rarr ._(7)N^(14)+beta^(c-)` The half`-` life period of `C^(14)` is 5770 year. The decay constant `(lambda)` can be calculated by using the following formuls `:` `lambda=(0.693)/(t_(1//2))` The comparison of the `beta^(c-)` activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method, however, ceases to be accurate over periods longer than 30000 years. The proportion of `C^(14)` to `C^(12)` in living matter is `1:10^(12)`. A nuclear explosion has taken place leading to an increase in the concentration of `C^(14)` in nearby areas. `C^(14)` concentration is `C_(1)` in nearby areas and `C_(2)` in areas far away. If the age of the fossil is determined to be `T_(1)` and `T_(2)` at the places , respectively, thenA. The age of the fossil will increase at the place where explosion has taken place and `T_(1)-T_(2)=(1)/(lambda)ln ``(C_(1))/(C_(2))`B. The age of the fossil will decrease at the place where explosion has taken place and `T_(1)-T_(2)=(1)/(lambda)ln``(C_(1))/(C_(2))`C. The age of fossile will be determined to be same.D. `(T_(1))/(T_(2))=(C_(1))/(C_(2))` |
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Answer» Correct Answer - A All radioactive decays are examples of first`-` order kinetics. So decay constant `lambda=(1)/(T_(1)-T_(2))ln``(C_(1))/(C_(2))` `C_(1)` is the concentration at `T_(1)` time. `C_(2)` is the concentration at `T_(2)` time. So, `T_(1)-T_(2)=(1)/(lambda)ln ``(C_(1))/(C_(2))` |
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| 148. |
Carbon-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)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1)` `.^(14)C` is absorbed by living organisms during photosynthesis. The `.^(14)C` content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of `.^(14)C` in the dead being, falls due to the decay which `.^(14)C` undergoes. `._(6)^(14)C rarr ._(7)^(14)N + beta^(-)` The half-life period of `.^(14)C` is 5770 years. The decay constant `(lamda)` can be calculated by using the following formula `lamda = (0.693)/(t_(1//2))`. The comparison of the `beta^(-)` activity of the dead matter with that of the carbon still in circulation enable measurement of the period of the 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)`. A nuclear explosion has taken place leading to increase in concentration of `.^(14)C` in nearby areas. `C^(14)` concentrations is `C_(1)` in nearby areas and `C_(2)` in areas far away. If the age of the fossil is determined to be `T_(1) and T_(2)` at the places respectively thenA. The age of the fossil will increase at the place where explosion has taken and `T_(1) - T_(2) = (1)/(lamda "ln" (C_(1))/(C_(2))`B. The age of the fossil will decrease at the place where explosion has taken place and `T_(1) - T_(2) = (1)/(lamda) "ln" (C_(1))/(C_(2))`C. The age of fossil will be determined to be sameD. `(T_(1))/(T_(2)) = (C_(1))/(C_(2))` |
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Answer» Correct Answer - A All radioactive decays are the examples of first order kinetics So decay constant `lamda = (1)/(T_(1) - T_(2)) "ln" (C_(1))/(C_(2))` `C_(1)` is the concentration at `T_(1)` times `C_(2)` is the concentration at `T_(2)` time So `T_(1) - T_(2) = (1)/(lamda) ln (C_(1))/(C_(2))` |
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| 149. |
Carbon-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)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1)` `.^(14)C` is absorbed by living organisms during photosynthesis. The `.^(14)C` content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of `.^(14)C` in the dead being, falls due to the decay which `.^(14)C` undergoes. `._(6)^(14)C rarr ._(7)^(14)N + beta^(-)` The half-life period of `.^(14)C` is 5770 years. The decay constant `(lamda)` can be calculated by using the following formula `lamda = (0.693)/(t_(1//2))`. The comparison of the `beta^(-)` activity of the dead matter with that of the carbon still in circulation enable measurement of the period of the 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 ageA. 6 yearsB. 6000 yearsC. 60,000 yearsD. It can be used to calculate any age |
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Answer» Correct Answer - B Radio carbon dating method ceases to be accurate over periods longer than 30,000 years |
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| 150. |
Carbon-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)^(14)N + ._(0)^(1)n rarr ._(6)^(14)C + ._(1)n^(1)` `.^(14)C` is absorbed by living organisms during photosynthesis. The `.^(14)C` content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of `.^(14)C` in the dead being, falls due to the decay which `.^(14)C` undergoes. `._(6)^(14)C rarr ._(7)^(14)N + beta^(-)` The half-life period of `.^(14)C` is 5770 years. The decay constant `(lamda)` can be calculated by using the following formula `lamda = (0.693)/(t_(1//2))`. The comparison of the `beta^(-)` activity of the dead matter with that of the carbon still in circulation enable measurement of the period of the 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 correctA. 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 organismsD. Carbon dating can not be used to determine concentration of `.^(14)C` in dead beings. |
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Answer» Correct Answer - C Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content as the ratio of `C^(14) " to " C^(12)` remains constant in living organism. |
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