InterviewSolution
Saved Bookmarks
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.
| 801. |
Choose the correct option (a) Atom bomb is based on nucleus fission (uncontrolled chain reaction)used as fissionable substance. In the exposion, a temperature of the order of `10^(6).^(@)C` and a pressure of lacs of atomspheres is obtained (ii) In nuclear reactor, by controlled chain reaction a huge enegry is obtained for useful purpose (iii) In a reactor fuel is `U^(234)` or `Pu^(238)` (iv) Moderators (Light water, heavy water, graphite, Cadmium rods are used as safety rodsA. `(i),(ii),(iii)`B. `(i),(ii),(iv)`C. `(ii),(iii),(iv)`D. all |
| Answer» Correct Answer - D | |
| 802. |
A nucleus with `Z = 92` emits the following in a sequence `a, beta^(bar) , beta^(bar)a,a,a,a,a, beta^(bar) , beta^(bar) , a, beta^(+) , beta^(+) , a ` Then `Z` of the resulting nucleus isA. 74B. 76C. 78D. 82 |
|
Answer» Correct Answer - C ( c) `Z_("Resulting nucleus") = 92 - 8 xx 2 + 4 xx 1 - 2 xx 1 = 78`. |
|
| 803. |
A radioactive element `._90 X^238` decay into `._83 Y^222`. The number of `beta-`particles emitted are.A. 4B. 6C. 2D. 1 |
|
Answer» Correct Answer - D (d) Number of `alpha`-particles emitted `= (238 - 222)/(4)` This decreases atomic number to `90 - 4 xx 2 = 82` Since atomic number of `._83 Y^222` is `83`, this is possible if one `beta`-particle is emitted. |
|
| 804. |
Choose the statement which is true.A. The energy released per unit mass is more in fission than in fusionB. The energy released per atom is more in fusion than in fission.C. The energy released per unit mass is more in fusion and that per atom is more in fission.D. Both fission and fusion produce same amount of enrgy per atom as well as per unit mass. |
|
Answer» Correct Answer - C The energy released per unit mass is more in fusion and that per atom is more in fission. |
|
| 805. |
In a uranium reactor whose thermal power is `P=100 MW`, if the average number of neutrons liberated in each nuclear spitting is `2.5`. Each splitting is assumed to release an energy `E=200 MeV`. The number of neutrons generated per unit time is-A. `4xx10^(18)s^(-1)`B. `8xx10^(19)s^(-1)`C. `8xx10^(19)s^(-1)`D. `(125)/(16)xx10^(18)s^(-1)` |
|
Answer» Correct Answer - D No. of nuclear spliting per second is `N=(100MW)/(200MeV)=(100)/(200xx1.6xx10^(-19))S^(-1)` No. of neutrons Liberated `=(100)/(200)xx(1)/(1.6xx10^(-19))xx2.5S^(-1)=(125)/(16)xx10^(18)S^(-1)` |
|
| 806. |
In a uranium reactor whose thermal power is `P=100 MW`, if the average number of neutrons liberated in each nuclear spitting is `2.5`. Each splitting is assumed to release an energy `E=200 MeV`. The number of neutrons generated per unit time is- |
|
Answer» Correct Answer - `[N = (v.P)/(E) = 0.8 xx 10^(19)s^(-1)]` |
|
| 807. |
Many unstable nuclie can decay spontaneously to a nucleus of lower mass but differnet combination of nucleons. The process of spontaneous emission of radiation is called radioactivity substance. Radioactive decay is a statistical process. Radiaactivity is independent of all external conditions. The number of decays per unit time or decay rate is called activity. Activity exponentially decrease with time. Mean lifetime is always greater than half-life time. Choose the correct statemnet about radioactivity:A. Radioactivity is a statistical process.B. Radioactivity is independent of high temperature and high pressureC. When a nucleus undergoes `alpha`- or - `beta`-decay, its atomic number changes.D. All of the these |
|
Answer» Correct Answer - d Radioactivity is indeopendent of all external condiations. When a nucleus undergoes an `alpha`-decay, its atomic number decreases by 2 and in beta decay, atomic number increases by `1`. |
|
| 808. |
Many unstable nuclie can decay spontaneously to a nucleus of lower mass but differnet combination of nucleons. The process of spontaneous emission of radiation is called radioactivity substance. Radioactive decay is a statistical process. Radiaactivity is independent of all external conditions. The number of decays per unit time or decay rate is called activity. Activity exponentially decrease with time. Mean lifetime is always greater than half-life time. If `T_(H)` is the half-life and `T_(M)` is the mean life. Which of the following statement is correct.A. `T_(M) gtT_(H)`B. `T_(M)ltT_(H)`C. Both ar directly proportional to square of the decay constant.D. `T_(M) prop lambda_(0)` |
|
Answer» Correct Answer - a `T_M =(1)/(lambda)` and `T_(H) =(0.693)/(lambda)` Now, `(T_(M))/(T_(H))=(1)/(lambda) xx (lambda)/(0.693)=(1)/(0.693)` i.e., `T_(M) gt T_(H)` From the above explanation, it is clear that choice (a) is correct and other (b), ( c) and (d) are wrong. |
|
| 809. |
A radioactive nuclie decay as follows `A overset(alpha)to A_(1)overset(beta) to A_(2) overset(alpha)to A_(3) overset(gamma)to A_(4)`. If mass number and atomic number of A are 180 and 72 respectively then mass number and atomic number for `A_4` will beA. 171 and 69B. 174 and 70C. 176 and 69D. 176 and 70 |
| Answer» Correct Answer - A | |
| 810. |
Two nuclie have mass numbers in the ratio `1:2`. What is the ration of their nuclear densities ? |
| Answer» Since the nuclear density is same for all nuclei. The ratio of their densities are `1:1 i.e. rho_(1):rho_(2)=1:1` | |
| 811. |
Two radioactive materials `X_1` and `X_2` have decay constants `10 lamda` and `lamda` respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of `X_1` to that of `X_2` will be `1//e` after a time.A. `lamda`B. `(1)/(2) lamda`C. `(1)/(4 lamda)`D. `( e)/(lamda)` |
|
Answer» Correct Answer - C ( c) If `N` is the number of radioactive nuclei present at some instant, then `N = N_0 e^(-lamda t)` The constant `N_0` represents the number of radioactive nuclei at `t = 0` Now, `(N_1)/(N_2) = (e^(-lamda_1 t))/(e^(-lamda_2 t))`or `(N_1)/(N_2) = (e^(-5 lamda t))/(e^(-lamda t)) = e^(-4 lamda t)` but `(N_1)/(N_2) = (1)/( e)` (as provided) Therefore, `(1)/( e) = (1)/(e^(4 lamda t))` or `4 lamda t = 1` or ` t = (1)/(4 lamda)`. |
|
| 812. |
In a radioactive material the activity at time `t_1` is `R_1` and at a later time `t_2`, it is `R_1`. If the decay constant of the material is `lamda`, thenA. `R_1 = R_2 e^(-lamda(t_1 - t_2))`B. `R_1 = R_2 e^(lamda(t_1 - t_2))`C. `R_1 = R_2(t_2//t_1)`D. `R_1 = R_2` |
|
Answer» Correct Answer - A (a) The decay rate `R` of a radioactive material is the number of decays per second From radioactive decay law, `-(dN)/(dt) prop N` or `- (dN)/(dt) = lamda N` Thus, `R = - (dN)/(dt)` or `R prop N` Or `R = lamda N` or `R lamda N_0 e^(-lamda t)`...(i) Where `R_0 = lamda N_0` is the activity of the radioactive material at time `t = 0` At time `t_1, R_1 = R_0 e^(-lamda t_1)` ...(ii) At `t_2, R_2 = R_0 e^(-lamda t_2)` ...(iii) Dividing Eq. (ii) by (iii), we have `(R_1)/(R_2) = (e^(-lamda t_1))/(e^(-lamda t_2)) = e^(-lamda(t_1 - t_2))` or `R_1 = R_2 e^(-lamda(t_1 -t_2))`. |
|
| 813. |
The average life `T` and the decay constant `lamda` of a radioactive nucleus are related asA. `T lamda = 1`B. `T =(0.693)/(lamda)`C. `(T)/(lamda) = 1`D. `T = ( c)/(lamda)` |
|
Answer» Correct Answer - A (a) Average life `T = ("Sum of all lives of all the atom")/("Total number of atoms") = (1)/(lamda)` `rArr T lambda = 1`. |
|
| 814. |
The radioactivity of a certain radioactive element drops to `1//64` of its initial value in `30` seconds. Its half-life is.A. 2 secondsB. 4 secondsC. 5 secondsD. 6 seconds |
|
Answer» Correct Answer - C ( c) `N = N_0 ((1)/(2))^(t//T) rArr (N_0)/(64) = N_0 ((1)/(2))^(30//T)` `rArr T = (30)/(6) = 5 sec`. |
|
| 815. |
The specific activity of radius is nearly-A. 1BqB. 1CiC. `3.7xx10^(10) Ci`D. 1 mCi |
| Answer» Correct Answer - B | |
| 816. |
A radioactive sample has `N_0` active at `t = 0` . If the rate of disintegration at any time is `R` and the number of atoms is `N`, them the ratio `R//N` varies with time as.A. .B. .C. .D. . |
|
Answer» Correct Answer - D (d) Rate `R = - (dN)/(dt) = lamda N_0 e^(-lamda t) = lamda N` `rArr ( R)/(N) = lamda (constant)` i.e., graph beween `( R)/(N)` and `t`, be a straight line parallel to the time axis. |
|
| 817. |
The graph between the instantaneous concentration (N) of a radioactive element and time (t) is.A. B. C. D. |
|
Answer» Correct Answer - D (d) By using `N = N_0 e^(-lambda t)` and `(dN)/(dt) = -lambda N` It shows that `N` decreases exponentially with time. |
|
| 818. |
The curve between the activity `A` of a radioactive sample and the number of active atoms `N` is.A. B. C. D. |
|
Answer» Correct Answer - B (b) `|(dN)/(dt)| = lamda N rarr |(dN)/(dt)| prop N`. |
|
| 819. |
If `20 gm` of a radioactive substance due to radioactive decay reduces to `10 gm` in `4` minutes, then in what time `80 gm` of the same substance will reduce to `10 gm` ?A. 8 minutesB. 12 minutesC. 16 minutesD. 20 minutes |
|
Answer» Correct Answer - B (b) `20 gm` substance reduces to `10 gm` (i.e., becomes half in `4 min`. So `T_(1//2) = 4 min`. Again `M = M_0 ((1)/(2))^(t//T_1//2)` `rArr 10 = 80((1)/(2))^(t//4) rArr (1)/(8) = ((1)/(2))^3 = ((1)/(2))^(t//4)` `rArr t = 12 min`. |
|
| 820. |
The example of nuclear fusion is.A. For of barium and krypton from unraniumB. Formation of helium from hydrogenC. Formation of plutonium `235` from uranium `235`D. Formation of water from hydrogen and oxygen. |
|
Answer» Correct Answer - B (b) `._1 H^2 + ._1 H^2 rarr ._2 He^4 + Q`. |
|
| 821. |
For a substance the average life for `alpha-`emission is `1620` years and for `beta-`emission is `405` years. After how much time the `1//4` of the material remains after `alpha` and `beta` emission ?A. 1500 yearsB. 300 yearsC. 449 yearsD. 810 years |
|
Answer» Correct Answer - C ( c) `lamda_alpha = (1)/(1620)` per year and `lamda_beta = (1)/(405)` per year and it is given that the fraction of the remained activity `(A)/(A_0) = (1)/(4)` Total decay constant `lamda = lamda_alpha +lamda_beta = (1)/(1620) + (1)/(405) = (1)/(324)` per year We know that `A = A_0 e^(- lamda t) rArr t = (1)/(lamda) log_e (A_0)/(A)` `rArr t = (1)/(lamda) log_e 4 - (2)/(lamda) log_e 2 = 324 xx 2 xx 0.693` =`449 years`. |
|
| 822. |
The rate of disintegration was observed to be `1017` disintegrations per sec when its half life period is `1445` years. The original number of particles are.A. `8.9 xx 10^27`B. `6.6 xx 10^27`C. `1.4 xx 10^16`D. `1.2 xx 10^17` |
|
Answer» Correct Answer - B (b) Rate of disintegration `(dN)/(dt) = 10^17 s^-1` Half life `T_(1//2) = 1445 year` =`1445 xx 365 xx 24 xx 60 xx 60 = 4.55 xx 10^10 sec` Now decay constant `lamda = (0.693)/(T_(1//2)) = (0.93)/(4.55 xx 10^10) = 1.5 xx 10^-11 per sec` The rate of disintegration `(dN)/(dt) = lamda xx N_0 rArr 10^17 = 1.5 xx 10^-11 xx N_0` `rArr N_0 = 6.6 xx 10^27`. |
|
| 823. |
A radio-isotope has a half-life of `5` yeard. The fraction of the atoms of this material that would decay in `15` years will beA. `1//8`B. `2//3`C. `7//8`D. `5//8` |
|
Answer» Correct Answer - C |
|
| 824. |
A ratio isotope has a half-life of `75` years. The fraction the atoms of this material that would decay in `150` years. Will be.A. `66.6 %`B. `85.5 %`C. `62.5 %`D. `75 %` |
|
Answer» Correct Answer - D (d) Number of half-lives in `150` years `n = (150)/(75) = 2` Fraction of the atom of decayed `1 - ((1)/(2))^n` =`1-((1)/(2))^2 = (3)/(4) = 0.75 rArr `Percentage decay `= 75 %`. |
|
| 825. |
The half-life of the isotope `._11 Na^24` is `15` hrs. How much time does it take for `(7)/(8) th` of a sample of this isotope to decay ?A. 75 hrsB. 65 hrsC. 55 hrsD. 45 hrs |
|
Answer» Correct Answer - D (d) Undecayed isotope `= 1 - (7)/(8) = (1)/(8)` `:. (N)/(N_0) = ((1)/(2))^(t//T) rArr ((1)/(8)) = ((1)/(2))^(t//15) rArr (t)/(15) = 3` or `t = 45 hours`. |
|
| 826. |
The sodium nucleus `._11^23 Na` contains.A. `11` electronsB. `12` protonsC. `23` protonsD. `12` neutrons |
|
Answer» Correct Answer - D (d) Number of neutrons `= A - Z = 23 - 11 = 12`. |
|
| 827. |
Regarding the nuclear forces, choose the correct options.A. They are short range focesB. They are charge independent forcesC. They are not electromagnetic forcesD. They are exchange forces |
|
Answer» Correct Answer - A::B::C::D from the property of nuclear fromces, it is answered. |
|
| 828. |
Regarding a nucleus choose the correct options.A. Density of a nucleus is directly proprtional to mass number `A`B. Density of all the nuclei is almost constant, of the order of `10^(17) kg//m^(3)`C. Nucleus radius is of the order of `10^(-15)m`D. Nucleus radius is directly proprtional to mass number, `A` |
|
Answer» Correct Answer - B::C density of all nuclei is almost equal and radius of the order of fermi. |
|
| 829. |
A nuclear raction is given us ` P+^(15)N rarr _(Z)^(A)X +n` (a). Find, `A,Z` and identity the nucleus `X`. (b) Find the `Q` value of the reaction . ( c) If the proton were to collide with the `.^(15)N` at rest, find the minimum KE needed by the proton to initiate the above reaction. (d) If the proton has twice energy in (c) and the outgoing neutron emerges at an angle of `90^(@)` with the direction of the incident proton, find the momentum of the protons and neutrons. `{:("[Given,"m(p)=1.007825 u","m(.^(15)C)=15.0106u","),(m(.^(16)N)=16.0061 u","m(.^(15)N)=15.000u"),"),(m(.^(16)O)=15.9949 u","m(u)=1.0086665u ","),(m(.^(15)O)=15.0031u"," and 1 u ~~ 931.5MeV."]"):}`. |
|
Answer» Correct Answer - (a) `Z=8,A =15 rArr X =._(8)O^(15)`, (b) `-3.67 MeV`, (c ) `3.9MeV`, (d) `P_(n)=79.4 MeV//c` (a) The nuleus is identify by:`Z=8,A=15 rArr X =_(8)O^(15)` (b) `Q=[m(P)+(m)(N^(15))-m(O^(15))-m(n)]c^(2)=-3.67 MeV` (c ) `K_(th)= -Q(1+(m_P)/(m_(M)))=3.9 MeV` (d) Now,`E_(k) =2 xx K_(th) =2 xx 3.9 MeV=7.8 MeV` and `Q = -3.63 MeV` (a) Conservation of momentum: `p_(0)cos theta =sqrt(2m_(p)E_(k))` `P_(0) sin theta =p_(n)` (b) Conservation of energy: `(p_(n)^(2))/(2m_(n))+(p_(0)^(2))/(2m_(0))=E_(k)+W` `p_(n)^(2)=(E_(k)(1-(m_(p))/(m_(0)))+Q)/((1)/(2m_(0))+(1)/(2m_(n)))` `:. P_(n) =79.4 MeV//c,P_(0)=145 MeV//C` and `theta =33^(@)` . |
|
| 830. |
`200 MeV` of energy may be obtained per fission of `U^235`. A reactor is generating `1000 kW` of power. The rate of nuclear fission in the reactor is.A. 1000B. `2xx10^(8)`C. `3.125xx10^(16)`D. 931 |
|
Answer» Correct Answer - C |
|
| 831. |
If the energy released in the fission of the nucleus is `200 MeV`. Then the number of nuclei required per second in a power plant of `6 kW` will be.A. `0.5 xx 10^14`B. `0.5 xx 10^12`C. `5 xx 10^12`D. `5 xx 10^14` |
|
Answer» Correct Answer - D (d) Energy released in the fission of one nucleus =`200 MeV` =`200 xx 10^6 xx 1.6 xx 10^-19 J = 3.2 xx 10^-11 J` `P = 16 KW = 16 xx 10^3 Watt` Now, number of nuclei required per second `n = (P)/( E) = (16 xx 10^3)/(3.2 xx 10^-11) = 5 xx 10^14`. |
|
| 832. |
To generate a power of `3.2` mega watt, the number of fissions of `U^235` per minute is. (Energy released per fission `= 200 MeV, 1 eV = 1.6 xx 10^-19 J`).A. `6 xx 10^18`B. `6 xx 10^17`C. `10^17`D. `6 xx 10^16` |
|
Answer» Correct Answer - A (a) Number of fission per second `=("Power output")/("Energy released per fission")` `=(3.2 xx 10^6)/(200 xx 10^6 xx 1.6 xx 10^-19) = 1 xx 10^17` rArr Number of fission per minute `=60 xx 1 xx 10^17 = 6 xx 10^18`. |
|
| 833. |
Assuming 1 metric ton of coal gives heat of combustion equal to 8 kcal and a single fission of `.^(235)U` releases 200 MeV. Calculate the minimum consumption in kg of `.^(235)U` to be heat equivalent ot 100 ton of coal |
|
Answer» Correct Answer - `[M = 4.077 xx 10^(-8) kg]` |
|
| 834. |
Calculate the energy released by `1 g` of natural uranium assuming `200 MeV` is released in eaech fission event and that the fissionable isotope `U^235` has an abundance of 0.7% by weight in natural uranium. |
|
Answer» Correct Answer - `[5.7 xx 10^(8) J]` |
|
| 835. |
If `200 MeV` energy is released in the fission of a single `U^235` nucleus, the number of fissions required per second to produce `1` kilowatt power shall be (Given `1 eV = 1.6 xx 10^-19 J`).A. `3.125xx10^(13)`B. `3.125xx10^(14)`C. `3.125xx10^(15)`D. `3.125xx10^(16)` |
|
Answer» Correct Answer - A |
|
| 836. |
The energy released by the fission of one uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 MW of power is :A. `10^(7)`B. `10^(10)`C. `10^(15)`D. `10^(17)` |
|
Answer» Correct Answer - D |
|
| 837. |
If `200 MeV` energy is released in the fission of a single `U^235` nucleus, the number of fissions required per second to produce `1` kilowatt power shall be (Given `1 eV = 1.6 xx 10^-19 J`). |
|
Answer» Energy released `=200 MeV` `=200xx10^(6)xx1.6xx10^(-19)=3.2xx10^(-11)J` `P=1` mega watt `=10^(6)` watts. `"No. of fission per second"(n)=("Total energy")/("Energy per fission")` `n=(10^(6))/(3.2xx10^(-11))=3.125xx10^(16)` Fissions |
|
| 838. |
`10^14` fissions per second are taking place in a nuclear reactor having efficiency `40%`. The energy released per fission is 250MeV. The power output of the reactor isA. 2000WB. 4000WC. 1600WD. 3200W |
|
Answer» Correct Answer - C Power output =0.4 x power input =0.4x 250 `xx10^(6) xx1.6xx10^(-19)xx10^(14)=1600 W`. |
|
| 839. |
There is a stream of neutrons with a kinetic energy of `0.0327 eV`. If the half-life of neutrons is `700 s`, what fraction of neutrons will decay before they travel is distance of `10 m`? Given mass of neutron `= 1.676 xx 10^(-27) kg` . |
|
Answer» Correct Answer - `0.004` Given that kinetic energy of neutrons is `(1)/(2) mv^(2)=0.0327 xx(1.6 xx10^(-19)) J` or `v^(2)=(2 xx 0.0327 xx(1.6 xx10^(-19)))/(1.675 xx10^(-27)) =625 xx10^(4)` or `v=2500 ms^(-1)` Time to travel a distance of 10 km is `(10^(4) (m))/(2500 m s^(-1))=4s` After `4s`, number of neutrons left can be given us `N =N_(0) 2^(-n)` where `n=(t)/(T)= no`. of half- lives. Here, `n=(4)/(700) =(1)/(175)` or `(N)/(N_(0)) =2^(-1//175)=0.996` or `N=0.996 N_0` Thus, fraction of neutrons decayed are `f=(N_(0)-N)/(N_(0)) =(0.004N_(0))/(N_(0)) =0.004` |
|
| 840. |
The energy liberated on complete fission of `1 kg` of `._92 U^235` is (Assume `200 MeV` energy is liberated on fission of `1` nucleus).A. `8.2 xx 10^10 J`B. `8.2 xx 10^9 J`C. `8.2 xx 10^13 J`D. `8.2 xx 10^16 J` |
|
Answer» Correct Answer - C ( c) Mass of a uranium nucleus =`92 xx 1.6725 xx 10^-17 + 147 xx 1.6747 xx 10^-27` =`393.35 xx 10^-27 kg` Number of nuclei in the given mass =`(1)/(393.35 xx 10^-17) = 2.542 xx 10^24` Energy released `= 200 xx 2.542 xx 10^24 MeV` =`5.08 xx 10^26 MeV = 8.135 xx 10^13 J ~~ 8.2 xx 10^13 J`. |
|
| 841. |
The energy liberated on complete fission of `1 kg` of `._92 U^235` is (Assume `200 MeV` energy is liberated on fission of `1` nucleus).A. `8.2xx10^(10) J`B. `8.2 xx10^(9) J`C. `8.2xx10^(13) J`D. `8.2xx10^(16) J` |
|
Answer» Correct Answer - C |
|
| 842. |
A nuclear reactor using `.^(235)U` generates `250 MW` of electric power. The efficiency of the reactor (i.e., efficiency of conversion of thermal energy into electrical energy) is `25%`. What is the amount of `.^(235)U` used in the reactor per year? The thermal energy released per fission of `.^(235)U` is `200 MeV`. |
|
Answer» Rate of electrical energy generration is `250 MW =250 xx 10^(6) W(or J s^(-1))`. So, electrical energy generation is `250 MW =250 xx 10^(6) W(or J s^(-1))`. Therefore, electrical energy generated in `1 year` is `(250 xx 10^(6) J s^(-1)) xx (365 xx 24 xx 60 xx 60 s)=7.884 xx 10^(15) J` Thermal eneregy from fission of one `.^(235)U` nuvleus is `200 MeV =200 xx 1.6 xx 10^(-13) = 3.2 xx 10^(-11J` . Since the efficiency is `25%`, the electrical energy obtained from the fission of one `.^(235)U` nucleus is `E_(1)=3.2 xx 10^(-11) xx(25)/(100) =8.0 xx 10^(-12) J` Therefore, the number of fission of `.^(235)U` required in one year will be `N=(7.884 xx10^(15))/(8.0 xx 10^(-12) =9.855 xx 10^(26))` Number of moles of `.^(235)U` required per year is `n=(9.855 xx 10^(26))/(6.023 xx 10^(23)) =1.636 xx10^(3)` Therefore, mass of `.^(235)U` required per year is `m=1.636 xx 10^(3) xx 235` `=3.844 xx 10^(5) g=384.4 kg`. |
|
| 843. |
Energy released in the fission of a single `._92 U^235` nucleus is `200 MeV`. The fission rate of a `._92 U^235` fuelled reactor operating at a power level of `5 W` is.A. `1.56 xx10^(-10) s^(-1)`B. `1.56 xx10^(11) s^(-1)`C. `1.56 xx10^(-16) s^(-1)`D. `1.56 xx10^(-17) s^(-1)` |
|
Answer» Correct Answer - B |
|
| 844. |
In a nuclear reactor an element `X` decays to a radio active element `Y` at a constant rate `10^(15)` atoms per sec. Each decay release `100 MeV` energy. Half life of `Y` equals `T` and decays to a stable product `Z`. Each decay of `Y` releases `50 MeV`. All energy released inside the reactor is used to produce electricity at an afficiency of `25%`. Calculate the electrical power in `kw` generated in the reactor in steady state. |
|
Answer» At steady state energy released per sec `etaxx r(E_(1)+E_(2)), eta=25% , r= 10^(15)` `E_(1)=100xx10^(6)xx1.6xx10^(-19)=1.6xx10^(-11)J` `E_(2)=50xx10^(6)xx1.6xx10^(-19)=0.8xx10^(-11)J` |
|
| 845. |
If the binding energy per nucleon in `._(3)Li^(7)` and `._(2)He^(4)` nuclei are respectively `5.60` MeV and `7.06` MeV, then the ebergy of proton in the reaction `._(3)Li^(7) +p rarr 2 ._(2)He^(4)` isA. `19.6` MeVB. `2.4` MeVC. `8.4` MeVD. `17.3` MeV |
|
Answer» Correct Answer - D |
|
| 846. |
If the binding energy per nucleon in `._(3)Li^(7)` and `._(2)He^(4)` nuclei are respectively `5.60` MeV and `7.06` MeV, then the ebergy of proton in the reaction `._(3)Li^(7) +p rarr 2 ._(2)He^(4)` isA. 19.6MeVB. 2.4 MeVC. 8.4MeVD. 17.3 MeV |
|
Answer» Correct Answer - D |
|
| 847. |
The binding energy per nucleon of `._(3)^(7) Li` and `._(2)^(4)He` nuclei are `5.60` MeV and `7.06` MeV, respectively. In the nuclear reaction `._(3)^(7)Li+._(1)^(1)H rarr ._(2)^(4)He+._(2)^(4)He+Q`, the value of energy `Q` released isA. `19.6` MeVB. `-2.4` MeVC. `8.4` MeVD. `17.3` MeV |
|
Answer» Correct Answer - D |
|
| 848. |
On an average, a neutron loses half of its energy per collision with a quesi-free proton. To reduce a `2 MeV` neutron to a thermal neutron having energy `0.04 eV`, the number of collisions requaired is nearly .A. `50`B. `52`C. `26`D. `15` |
|
Answer» Correct Answer - c Let n collisions are required for the given conditiaon. Then, `((1)/(2))^(n) xx 2 MeV=0.04 xx 10^(-6) MeV` `2^(n) =(2)/(0.04 xx 10^(6)) =50xx10^(6)` After solving above equation, `n=26. |
|
| 849. |
If the binding energy per nucleon in `._(3)Li^(7)` and `._(2)He^(4)` nuclei are respectively `5.60` MeV and `7.06` MeV, then the ebergy of proton in the reaction `._(3)Li^(7) +p rarr 2 ._(2)He^(4)` isA. 19.60 MeVB. 12.26 MeVC. 8.46 MeVD. 17.28Mev |
|
Answer» Correct Answer - D |
|
| 850. |
A neutron makes a head-on elastic collision with a stationary deuteron. The fraction energy loss of the neutron in the collision is |
|
Answer» Correct Answer - [(a) `eta = 4Mm (m+m)^(2) = 0.89`, (b) `eta = (2m)/((m+M)) = (2)/(3)`] |
|