InterviewSolution
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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.
| 1. |
The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is ……..A. `2.2MeV`B. `23.6MeV`C. `28.0MeV`D. `30.2MeV` |
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Answer» Correct Answer - B `._(1)H^(2) +._(1)H^(2)rarr _(2)He^(4)` Energy `= 4xx7 - 4xx1.1 = 23.6MeV` |
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| 2. |
The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is ……..A. 20.8 MeVB. 16.6MeVC. 25.2MeVD. 23.6MeV |
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Answer» Correct Answer - d Energy released would be `Delta E` =total energy of `._(2)He^(4)` `- 2xx` (total binding energy of `._(1)_(He)^(4)` ` 4 xx 7.0 -2(1.1)(2)` `=23.6 MeV` . |
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| 3. |
The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is …….. |
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Answer» `._(1)^(2)H+._(1)^(2)H rarr._(2)^(4)He` Initial `B.E. = 2[2xx1.1] = 4.4 MeV` Final `B.E. = 4xx 7 = 28 MeV` Final `B.E. gt` Initial `B.E.` Energy released `= 28 - 4.4 = 23.6 MeV` |
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| 4. |
A neutron breaks into a proton and electorn. Calculate the eenrgy produced in this reaction in `m_(e) = 9 xx 10^(-31) kg, m_(p) = 1.6725 xx 10^(-27) kg, m_(n) = 1.6747 xx 10^(-27) kg, c = 3xx10^(8)m//sec`. |
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Answer» `._0^(1)nrarr_(1)^(1)p+ ._(-1)^(0)e` Mass defect: `Deltam = [1.6747 xx 10^(-27) - 1.3725 xx 10^(-27)-9xx10^(-31)]` `= 1.3 xx 10^(-30) kg` Energy produced `= Deltamc^(2) = 1.3 xx 10^(-30) xx (3xx10^(8))^(2)` `= 11.7xx10^(-14) J = (11.7 xx 10^(-14))/(1.6xx10^(-13)) = 0.73 MeV` |
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| 5. |
`A overset(lambda_(1)) rarr B overset (lambda_(2))rarr C` t=0 `N_(0)` 0 0 ` t N_(1) N_(2) N_(3)` In the above radioactive decay C is stable nucleus. Then:A. rate of decay of `A` will first increase and then decreasesB. number of nuclei of `B` first increase and then decreasesC. if `lambda_(2) gt lambda_(1)`, then activity of `B` will always be higher than activity of `A`D. if `lambda gt gt lambda_(2)`, then number of nucleus of `C` will always be less than number of nucleus of `B`. |
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Answer» Correct Answer - B Rate of decay of `A` keeps on decreasing continuosly because concentration of `A` decreases with time. `rArr A` is false Initial rate fo production of `B` is `lambda_(1)N_(0)` and rate of decay is zero. With time, as the number of `B` atom in increase, the rate of its production decreases and its rate of decay increases. Thus the number of nuclei of `B` will first increase and then decrease. `rArr B` is the correct choice The initial activity of `B` zero whereas initial activity of `A` is `lambda_(1)N_(0) rArrC` is false. As time `t rarr : N_(A)=0,N_(B)=0` and `N_( C)=N_(0)rArrD` is false |
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| 6. |
`A overset(lambda_(1)) rarr B overset (lambda_(2))rarr C` t=0 `N_(0)` 0 0 ` t N_(1) N_(2) N_(3)` In the above radioactive decay C is stable nucleus. Then:A. rate of decay of A will first increase and then decreaseB. Number of nuclei of B will first increase and then decreaseC. if `lambda_(2) gt lambda_(1)` then activity of B will always be higher than activity of AD. If `lambda_(1) gt gt lambda_(2)` then number of nucleus of C will always be less than number of nucleus of B. |
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Answer» Correct Answer - B |
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| 7. |
A sample has `4 xx 10^16` radioactive nuclei of half-life `10` days. The number of atoms decaying in `30` days is.A. `3.9 x 10^16`B. `5 xx 10^15`C. `10^16`D. `3.5 xx 10^16` |
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Answer» Correct Answer - D (d) `N = 4 xx 10^16 ((1)/(2))^((30)/(10)) = (1)/(2) xx 10^16` Atoms decayed `= 4 xx 10^16 - (1)/(2) xx 10^16` =`3.5 xx 10^16`. |
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| 8. |
(a) Assume that a particular nucleus, in a radioactive sample, has a probability `rho` of decaying in the interval t to `t+Deltat`. Taking that the nucleus actually does not decay in the above interval, what is probability that it will decay in the interval t `+Deltat` to `t+2Deltat`? (b) Find the mean life of `.^(55)C_(0)` is its activity is known to decreases 4% per hour. `ln((25)/(24))=0.04`. |
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Answer» Correct Answer - (a) p (b) 25hr |
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| 9. |
The decay constant `lambda` of a radioactive sample:A. Decreases with increase of external pressure and temperatureB. Increase with increase of external pressure and temperatureC. Decreases with increase of temperature and increase of pressureD. Is independent of temperature and pressure. |
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Answer» Correct Answer - D |
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| 10. |
In a sample of radioactive material the probability that a particular nucleus will decay in next 2 hour is `8xx10^(-4)` find the half life of the radioactive sample. [Take `ln2=0.693,ln(0.9992)=-8xx10^(-4)`] |
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Answer» Correct Answer - 1732.5hr |
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| 11. |
A radioactive sample has half-life of `5` years. Probability of decay in `10` years will be.A. `50%`B. `75%`C. `60%`D. `100%` |
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Answer» Correct Answer - B |
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| 12. |
Calculate the time taken to decay `100` percent of a radioactive sample in terms of (a) half- life T and (b) mean-life `T_(av)` |
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Answer» We know that `N=N_(0) e^(_lambda t)`, where `lambda =((In 2)/(T)) =(1)/(T_(av))` Here, `N=(1-eta)N_(0)` ` (1-eta)N_(0)=N_(0) e^(-lambda t)` `t=(1)/(lambda)1n|(1)/((1-eta))|` (a) Thus, `t=(T)/(1n2)[1n|(1)/(1-eta)|]=(T)/(0693)In|(1)/(1- eta)|` `t=T_(av) In|(1)/(1-eta)|`. |
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| 13. |
A radioactive sample has half-life of `5` years. Probability of decay in `10` years will be.A. 1B. 0.75C. 0.5D. 0.25 |
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Answer» Correct Answer - B (b) Number of half-lives `n = (10)/(5) = 2`, Now `(N)/(N_0) = ((1)/(2))^2 = (1)/(4)` Fraction decayed `=1 - (N)/(N_0) = 1-(1)/(4) = (3)/(4)` ` rArr` In percentage `= (3)/(4) xx 100 = 75 %`. |
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| 14. |
If half-life of a substance is `3.8` days and its quantity is `10.38 gm`. Then substance quantity remaining left after `19` days will beA. `0.151 gm`B. `0.32 gm`C. `1.51 gm`D. `0.16 gm` |
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Answer» Correct Answer - B (b) Number of half-lives `n = (19)/(3.8) = 5`, Now `(N)/(N_0) = ((1)/(2))^n` `rArr (N)/(10.38) = ((1)/(2))^5 rArr N = 10.38 xx ((1)/(2))^5 = 0.32 gm`. |
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| 15. |
Two radioactive materials `X_(1)` and `X_(2)` have decay constant `11 lambda` and `lambda` respectively. If initially they have same number of nuclei, then ratio of number of nuclei of `X_(1)` to `X_(2)` will be `(1)/(e^(2))` after a timeA. `(1)/(5 lambda)`B. `(1)/(11 lambda)`C. `(1)/(10 lambda)`D. `(1)/(9 lambda)` |
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Answer» Correct Answer - A `N_(1)=N_(o)e^(-11 lambda t)` and `N_(2) = N_(0) e^(-lambdat)` It is given that `(N_(1))/(N_(2))=(1)/(e^(2))`, solving the above two equations, we get `t=(1)/(5 lambda)` |
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| 16. |
The Process by which a heavy nucleus splits into light nuclei is known as-A. fissionB. `alpha`-decayC. FussionD. chain reaction |
| Answer» Correct Answer - A | |
| 17. |
Suppose, the daughter nucleus in a nuclear decay is itself radioactive. Let `lambda_p and lambda_d` be the decay constants of the parent and the daughter nuclei. Also, let `N_p and N_d` be the number of parent and daughter nuclei at time `t`. Find the condition for which the number of daughter nuclei becomes constant. |
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Answer» Correct Answer - `[lambda_(p)N_(p) = lambda_(d)N_(d)]` |
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| 18. |
A nucleus splits into two nuclear parts having radii in the ratio `1:2` Their velocities are in the ratioA. `8:1`B. `6:1`C. `4:1`D. `2:1` |
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Answer» Correct Answer - A `m_(1)v_(1), (m_(1))/(m_(2))=((R_(1))/(R_(2)))^(3)=(V_(2))/(V_(1))` |
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| 19. |
A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to `2 : 1`. What will be the ratio of their nuclear size (nuclear radius)?A. `2^(1//3):1`B. `1:2^(2//3)`C. `3^(1//2):1`D. `1:3^(1//2)` |
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Answer» Correct Answer - B |
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| 20. |
A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to `2 : 1`. What will be the ratio of their nuclear size (nuclear radius)?A. `2^(1//3) : 1`B. `1 : 2^(1//3)`C. `3^(1//2) : 1`D. `1 : 3^(1//2)` |
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Answer» Correct Answer - B |
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| 21. |
The nuclear size is measured in units ofA. AngstromB. FermiC. BarD. Light-year |
| Answer» Correct Answer - B | |
| 22. |
A radioactive substance is being consumed at a constant of `1 s^(-1)`. After what time will the number of radioactive nuclei becoem `100`. Initially, there were 200 nuceli present.A. 1 secB. `1/(l n (2))` secC. In (2) secD. 2 sec |
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Answer» Correct Answer - C |
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| 23. |
The energy released by the fission of a single uranium nucleus is `200 MeV`. The number of fission of uranium nucleus per second required to produce `16 MW` of power is (Assume efficiency of the reactor is `50%`)A. `2xx10^(6)`B. `2.5xx10^(6)`C. `5xx10^(6)`D. `1xx10^(18)` |
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Answer» Correct Answer - D No of fission = total energy produced/(energy released per fission `xx` efficiency) |
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| 24. |
A radioactive substance is being consumed at a constant of `1 s^(-1)`. After what time will the number of radioactive nuclei becoem `100`. Initially, there were 200 nuceli present.A. `1 s`B. `(1)/(1 n(2))s`C. `ln(2)s`D. `2s` |
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Answer» Correct Answer - c Let N be the number of nuclei at any time t. Then, `(dN)/(dt) =200 -lambda N` `:.int_(0)^(N) (dN)/((200 -lambda N)) = int_(0)^(t)dt` or N=(200)/(lambda) (1-e^(-lambda t))` Given: `N =100 and lambda =1 s^(-1)` `:. 100=200 (1-e^(-t))` or `e^(-t)=((1)/(2)) :. t=In(2) s`. |
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| 25. |
A radiaocatice isotope is being produced at a constant rate `X`. Half-life of the radioactive substance is `Y`. After some time, the number of radioactive nuceli become constant. The value of this constant is .A. `(XY)/(ln(2))`B. `XY`C. `(XY)ln(2)`D. `(X)/(Y)` |
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Answer» Correct Answer - a Number of radio nulei become constant, when rate of production becoms equal to rate of decay. `X=lambda N` or `N=(X)/(lambda)` Given, `y = In(2)/(lambda)` `rArr N=(XY)/(1n(2))`. |
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| 26. |
(a) Derive the law of radioavtive decay, `viz`. `N=N_(0)e^(-lambda t)` (b) Explain, giving necessary reactions, how energy is released during (i) fission and (ii) fusion. |
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Answer» (a) The following laws, known as the laws of radioactive decay: (1) It is a spontaneous phenomenon and one cannot predict, when a particular atom will undergo disintergration. According to radioactive decay law, `(dN)/(dt)propN " "(dN)/(dt)=lambdaN`.....(i) From the equation (i) `(dN)/(dt)= -lambdadt` Intergrating, we have `int (dN)/(N)= -lambdaint dt` or `Log_(e )N= -lambdat+K`, ......(ii) Where `K` is constant of intergration, when `t=0, N=N_(0)` On Setting `t=0` and `N=N_(0)` the equation (ii) `log_(e ) N_(0)= -lambdaxx0+k` `k=log_(e )N_(0)` Substituting for `K` in the equation (ii) `log_(e )N= -lambdat+log_(e )N_(0)` `"log"_(e )(N)/(N_(0))-lambda t , (N)/(N_(0))-e^(-lambdat)` `N=N_(0)e^(-lambda t)` (b) (i) The fission reaction of `._(92)uu^(325)` may be represented as given below: `{:(._(92).^(235),._(0)n^(1),[._(92).^(236)],),(._(56)Ba^(141),._(36)Kr^(92),3_(0)n^(1),Q):}` The energy `(Q)` released was estimated to be `200 MeV` per fission (or about `0.9 MeV` per nucleon) and is equivalent to the diffecence in masses of the nuclei before and after the fission. (ii) When two or more than two light nuclei fuse together to form harry nucleus wiht the liberation of energy, the releasing `24 MeV` of energy. The fusion reaction may be expressed as follow: ` H^(2)+_(1)H^(2)rarr_(2)H^(4)e+24MeV` The above nuclear fusion reaction is energetically possible, only if the mass of the `_(2)H^(4)e` nucleus is less than the sum of the massess of wo deuteron nuclei. |
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| 27. |
The acticity of a sample is `64 xx 10^-5 Ci`. Its half-life is `3` days. The activity will become `5 xx 10^-6 C i` after.A. 12 daysB. 7 daysC. 18 daysD. 21 days |
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Answer» Correct Answer - D (d) `A = A_0 ((1)/(2))^(t//T_1//2) rArr 5 xx 10^-6 = 64 xx 10^-5 ((1)/(2))^(t//3)` `(1)/(128) = ((1)/(2))^(t//3) rArr t = 21 days`. |
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| 28. |
`A overset(lambda)rarr B overset(2 lambda)rarr C` `T=0 ,N_(0) , 0 `, ` T N_(1) N_(2) N_(3)` The ratio of `N_(1)" to "N_(2) ` is maximum I sA. at no time this is possibleB. ``2C. `1//2`D. `(ln2)/(2)` |
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Answer» Correct Answer - b `(dN_(2))/(dt)=lambda N_(1) -2 lambda N_(2)` For `n_(2)` to be maximum, `(dN_(2))/(dt)=0` `rArr lambda N_(1)= 2 lambda N_(2)` or `(N_(1))/(N_(2)) =2`. |
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| 29. |
A radioactive element` X` converts into another stable elemnet `Y`. Half-life of `X` is `2h`. Initally, only `X` is present. After time `t`, the ratio of atoms of `X` and `Y` is found to be `1:4` Then `t` in hours is .A. `2`B. `4`C. between` 4` and `6`D. `6` |
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Answer» Correct Answer - c Let `N_(2)` be the number of atoms of X at time `t =0`. Then, at `t=4h` (two half-lives), `N_x =N_0/4` and `N_y =3N_0/4` `because (N_(x))/(N_(y)) =(1)/(3) ~~0.33` At `t=6th` (three half-lives), `N_(x) =(N_(0))/(8)` and `N_(y) =(7N_(0))/(8)` or `(N_(x))/(N_(y)) =(1)/(7) ~~0.142` The given ratio `(1)/(4)` lies between `(1)/(3)` and `(1)/(7)`. Therefore , t lies between `4 h` and `6h`. |
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| 30. |
The fossil bone has a `.^(14)C` : `.^(12)C` ratio, which is `[(1)/(16)]` of that in a living animal bone. If the half -life of `.^(14)C` is `5730` years, then the age of the fossil bone is :A. 11460 yearsB. 17190 yearsC. 45840 yearsD. 22921 years |
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Answer» Correct Answer - D (d) After `n` half-lives (`Le`., at `t = nT`) the number of nuclides left undecayed, `N = N_0 ((1)/(2))^n` Given, `(N)/(N_0) = (1)/(16)` `:. (1)/(16) = ((1)/(2))^n` or `((1)/(2))^4 = ((1)/(2))^n` Equating the powers, we obtain `n= 4` i.e., `(t)/(T) = 4` or `t = 4 T` or `t = 4 xx 5730 = 22920 years` `(because T = 5730 years)`. |
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| 31. |
The isotope `._6^(14)C` is radioactive and has a half-life of 5730 years . If you starts with a sample of `1000` carbon `-14` niclei, how many will still be around in `17,190 years`? |
| Answer» In `5730 years` half the sample will have decayed, leaving 500 radioactive `._(6)^(14)C` nuclei . In another `5730 year`s (for a total elapsed time of `11460 years`), the number will be reduced to `250` nuclei reamin. These numbers represent ideal circumstances. Radioactive decay is an averging process over a very large number of atoms, and the actual outcome depends on statistic. Our original sample in this example contained only `1000` nuclei, certainly not a very large number, Thus, if we were actullaly to count the number remaining after one half -life for this small sample, it probably would not be exactly `500`. | |
| 32. |
Find the energy liberated in the beta decay of `._6^(14)C " to " ._7^(14)N` as represented by Eq.(iii). Equation (iii) refers to nuclei. Adding six electrons to both sides of Eq.(iii) gives ` .-6^(14)C "atom" rarr .-7^(14)N`. |
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Answer» We know that `._(6)^(14)C` has a mass of `14.003 074u` . Here, the mass differrence between the initial and final states is `Delta m=14.003 242 u -14.003 74 u = 0.000 168 u` This corresponds to an energy release of `E = (0.000 168 u)(931 . 494 MeV//u)=0.156 MeV`. |
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| 33. |
The binding energy of `._(17)^(35 )Cl` nncleus. Take atomic mass of `._6^(12) C` as `12.000 an u` Take `R_(0) =1.2 xx 10-^(15) m`. |
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Answer» Correct Answer - `34.969 MeV` The `._(17)^(35)C1` nucleus has `17` protons and `18` neutron s . Therefore, the mass of contents nucleus of `._(17)^(35)C1` is `M=17 m_(p) +18m_(n)=17 xx1.007825 +18 xx 1.0086645` `35.289 a.m.u.` Now, mass defect for the nucleus is `Delta m=(298 MeV)/(9312 MeV//a.m.u.) =0.3200 a.m.u.` Thus, atomic mass of `._(17)^(35)C1` = mass of contents nucleus -mass defect `m-Delta m=35.289 a.m.u. -0.3200 a.m.u.` `=34.969 a.m.u.` |
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| 34. |
Find the binding energy of the nucleus of lighium isotope `._(3)Li^(7)` and hence find the binding energy per nucleon in it. |
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Answer» Correct Answer - `B.E=[3M_(1H^(1))+4m_(0n^(1))-M_(3Li^(7))]931 MeV` `=39.22 MeV, (B.E)/(A)=(39.22)/(7)=5.6 MeV` `B.E=[3M_(1H^(1))+4m_(0n^(1))-M_(3Li^(7))]931 MeV=39.22MeV, (B.E.)/(A)=(39.22)/(7)=5.6 MeV` |
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| 35. |
Find the density of `._(6)^(12)C` nucleus. Take atomic mass of `._(6)^(12)C` as `12.00 am u` Take `R_(0) =1.2 xx10^(-15) m` . |
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Answer» Correct Answer - `2.4 xx 10^(7) kgm^(-3)` The radius of `._(6)^(12)C` is `12 a.m.u.` Neglecting the masses and binding energies of the six electrons. Nuclear density `=(m)/((4)/(3)pi R^(3)) =(12 xx 1.66 xx 10^(-27))/(((4)/(3) pi)(2.7 xx 10^(-15))^(3))` `=2.4 xx 10^(17)` kg m^(-3)` . |
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| 36. |
Assuming the radius of a nucleus to be equal to `R=1.3 A^(1//3)xx10^(-15)m`. Where `A` is its mass number, evaluate the density of nuclei and the number of nucleons per unit volume of the nucleus. Take mass of one nucleon `=1.67xx10^(-27)kg` |
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Answer» Correct Answer - `2xx10^(11)kg//cm^(3), 1xx10^(38)n ucl//cm^(3)` The radius of nucleus is `R=R_(0)A^(1//3)` where `A=` mass number `R_(0)=1.3xx10^(-15)m=1.3 fm` The volume of nuclues is `=(4)/(3)piR^(3)=(4)/(3)pi(R_(0)A^(1//3))=(4)/(3)piR_(0)^(3)A` A`=` mass number `=` number of necleons `:.` The number of nucleons per unit volume is `=(A)/(V)` `(1)/((4)/(3)piR_(0)^(3))=1.09xx10^(38)"necleons"//"cc"` Density is " "`rho=("mass")/("volume")` `=1.09xx10^(38)xx"mass of nucleons per cc"` `=1.82xx10^(11)kg//"cc"` |
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| 37. |
A neutron star has a density equal to that of the nuclear matter. Assuming the staar to be spherical, find the radius of a neutron star whose mass is `4.0xx10^30`kg (twice the mass of the sun ). |
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Answer» Correct Answer - `r_(1)=[(4xx10^(30))/(3xx10^(17))xx(1)/(4pi)]^(1//3)= 14.71 km` (ii) `r_(2)=[(6xx10^(24))/(3xx10^(17))xx(3)/(4pi)]^(1//3)=168.4m` |
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| 38. |
A mathc box of `5 cmxx5cmxx1cm` dimensions is filled with nuclear matter. Its weight is in the order ofA. `10 g`B. `10^(8) g`C. `10^(12) g`D. `10^(15) g` |
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Answer» Correct Answer - D `D=(M)/(V)` |
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| 39. |
Fusion processes, like combining two deuterons to form a `He` nucleus are impossible at ordinary temperature and pressure. The reasons for this can be traced to the fact:A. nuclear forces have short range.B. nuclei are positively charged.C. the original nuclei must be completely ionized before fusion can take place.D. the original nuclei must first break up before combining with each other. |
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Answer» Correct Answer - A::B To fuse nuclei, their must be very small from each other. At ordinary temperature and pressure, coulomb repulsine forces between the nuclei dominate over nuclear forces. (since the latter is short ranged) |
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| 40. |
Among the following reactions which is impossibleA. `._(2)He^(4)+._(4)Be^(9)=._(0)n^(1)+._(6)C^(12)`B. `._(2)He^(4)+N^(14)=._(1)H^(1)+._(8)O^(17)`C. `4(._(1)H^(1))=._(2)He^(4)+2(._(1)e^(0))`D. `._(3)Li^(7)+._(1)H^(1)=._(4)Be^(8)` |
| Answer» Correct Answer - C | |
| 41. |
In an endo-ergic reaction the binding energies of reactants and products are `E_(1), E_(2)` respectivelyA. `E_(1) lt E_(2)`B. `E_(1)=E_(2)`C. `E_(1) gt E_(2)`D. `E_(1) gt E_(2)` |
| Answer» Correct Answer - C | |
| 42. |
`1 kg` of iron (specific heat `120 Cal Kg^(-1) C^(-1)`) is heated by `1000^(@)C`. The increases in its mass isA. zeroB. `5.6xx10^(-8) Kg`C. `5.6xx10^(-16)Kg`D. `5.6xx10^(-12) Kg` |
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Answer» Correct Answer - D `Delta mc^(2)=J(MS Delta t)` |
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| 43. |
The binding energies of the atom of elements `P` and `Q` are `E_(P)` and `E_(Q)` respectively. There atoms of element `Q` fuse on atom of element `P`. The correct relation between `E_(P), E_(Q)` and `e` will beA. `E_(1)-3E_(b)=Q`B. `3E_(b)-E_(1)=Q`C. `E_(1)+3E_(b)=Q`D. `E_(b)+3E_(b)=Q` |
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Answer» Correct Answer - A `3E_(b)rarrE_(a)+Q` |
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| 44. |
The binding energies per nucleon for deuterium and helium are `1.1 MeV` and `7.0 MeV` respectively. What energy in joules will be liberated when `2` deuterons take part in the reaction.A. `18.88xx10^(-3) J`B. `18.88xx10^(-5) J`C. `18.88xx10^(-7) J`D. `18.88xx10^(-10) J` |
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Answer» Correct Answer - C `E=4((B.E)/(A) "of" He-(B.E)/(A)"of Deuterium")` |
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| 45. |
The binding energies per nucleon for deuterium and helium are `1.1 MeV` and `7.0 MeV` respectively. What energy in joules will be liberated when `2` deuterons take part in the reaction. |
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Answer» `._(1)^(2)H + ._(1)^(2)H rarr ._(2)^(4)He + Q` Binding energy per nucleon of helium `(._(2)^(4)He) = 7 MeV` Binding energy `= 4 xx 7 =28 Me V` Binding energy nucleon of deuterium `(._(2)^(1)H) = 1.1 MeV` Binding energy `= 2 xx 1.1 = 2.2 MeV` Energy liberated `(Q) = (28 - (2.2)2] = 23.6 MeV` i.e., `Q = 23.6 xx 10^(6) xx 6 xx 10^(-19), Q = 37.76 xx 10^(-13)J` |
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| 46. |
A radioactive with decay constant `lambda` is being produced in a nuclear ractor at a rate `q_(0)` per second, where `q_(0)` is a positive constant and t is the time. During each decay, `E_(0)` energy is released. The production of radionuclide starts at time `t=0`. Which differential equation correctly represents the above process?.A. `(dN)/(dt)+lambda N=q_(0) t`B. `(dN)/(dt)-lambda N=q_(0) t`C. `(dN)/(dt)+q_(0) t=lambda N`D. `(dN)/(dt)+q_(0) t=-lambda N` |
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Answer» Correct Answer - a `(dN)/(dt)=q_(0) t-lambda N,(dN)/(dt)+lambda N=q_(0) t` . |
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| 47. |
When two deuterium nuclei fuse together to form a tritium nuclei, we get aA. neutronB. deuteronC. alpha particleD. proton |
| Answer» Correct Answer - D | |
| 48. |
A radioactive with decay constant `lambda` is being produced in a nuclear ractor at a rate `q_(0)` per second, where `q_(0)` is a positive constant and t is the time. During each decay, `E_(0)` energy is released. The production of radionuclide starts at time `t=0`. Which differential equation correctly represents the above process?.A. `(dN)/(dt)+a_(0)t=lambda N`B. `(dN)/(dt)-lambda N=a_(0)t`C. `(dN)/(dt)+lambda N=a_(0)t`D. `(dN)/(dt)+a_(0)t= - lambda N` |
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Answer» Correct Answer - C `(dN)/(dt)=a_(0)t-lambdaN, (dN)/(dt)+lambda N=a_(0)t` |
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| 49. |
A deuterium reaction that occurs in an experimental fusion reactor is in two stage: (a) Two deuterium `(._1^2D)` nuclei fuse together to form a tritium nucleus, with a proton as a by product written as `D(D,p)T`. (b) A tritium nucleus fuses with another deuterium nucleus to form a helium `._2^4He` nucleus with neutron as a by - product, written as T (D,n) `._2^4He`. Compute (a) the energy released in each of the two stages, (b) the energy released in the combined reaction per deutrium. (c ) What percentage of the mass energy of the initial deuterium is released. Given, `{:(._1^2D=2.014102 am u),(._1^3 T=3.016049), (._2^4 He =4.002603 am u),(._1^1 H =1.007825 am u),(._0^1 n =1.00665 am u):}`.A. `0.5%`B. `0.7%`C. `0.38%`D. `1.18%` |
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Answer» Correct Answer - D Mass defect concept |
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| 50. |
(a) Find the energy needed to remove a neutron from the nucleus of the calcium isotopes `._(20)^(42)Ca` (b) Find the energy needed to remove a proton from this nucleus (c ) Why are these energies different ? Atomic masses of `._(20)^(41)Ca` and `._(20)^(42)Ca` are `40.962278 u` and `41.958622 u` respectively. |
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Answer» (a) Removing a neutron from `._(20)^(42)Ca` leaves `._(20)^(41)Ca`. The mass of `._(20)^(41)Ca` plus the mass of a free neutron is: `40.962278 u +1.008665 u =41.970943 u` The difference between this mass and the mass of `._(20)^(42)Ca` is `0.012321 u` , so the binding energy of the mission neutron is: `(0.012321 u) (931.49 MeV//u)=11.48 MeV` (b) Removing a proton from `._(20)^(42)Ca` leaves the potassium isotope `._(19)^(41)K`. A similar calculation gives a binding energy of `10.27 MeV` for the missing proton. (c ) The neutron was acted upon only by attractive nuclear whereas the proton was also acted upon by repulsive electric forces that decrease its binding energy. |
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