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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. |
A nucleus X, initially at rest , undergoes alpha dacay according to the equation , ` _(92)^(A) X rarr _(Z)^(228)Y + alpha ` (a) Find the value of `A` and `Z` in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius `0.11 m` in a uniform magnetic field of `3` Tesla find the energy (in MeV) released during the process and the binding energy of the parent nucleus X Given that `: m (Y) = 228.03 u, m(_(0)^(1)n) = 1.0029 u. ` `m (_(2)^(4) He) = 4.003 u , m (_(1)^(1) H) = 1.008 u `A. `2.5`B. `4.7`C. `9.9`D. 8 |
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Answer» Correct Answer - B `m_(y)v_(y)=m_(alpha)v_(alpha)rArrV_(y)~~2.115xx10^(5)m//s` `:.` TE released during an a decay of the nucleus X is. |
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| 52. |
A nucleus X, initially at rest , undergoes alpha dacay according to the equation , ` _(92)^(A) X rarr _(Z)^(228)Y + alpha ` (a) Find the value of `A` and `Z` in the above process. (b) The alpha particle produced in the above process is found to move in a circular track of radius `0.11 m` in a uniform magnetic field of `3` Tesla find the energy (in MeV) released during the process and the binding energy of the parent nucleus X Given that `: m (Y) = 228.03 u, m(_(0)^(1)n) = 1.0029 u. ` `m (_(2)^(4) He) = 4.003 u , m (_(1)^(1) H) = 1.008 u `A. `2.3`B. `4.7`C. 6D. `7.8` |
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Answer» Correct Answer - D Mass lost during a-decay, `Deltam=(E)/(931)=0.005U` `:.` Mass of nucleus X, `m_(x)=m_(y)+m_(alpha)+Deltam=225.038 U` `:.` mass defect in nucleus X, `m_(d) = {[92m_(p)+(225-92)m_(n)]-m_(x)}U=1.895 U` `:. BE` per nucleon in nucleus `X= (m_(d)xx931)/(225)=7.84 MeV` |
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| 53. |
The half - life of `^(215) At` is `100 mu , s.`The time taken for the radioactivity of a sample of `^(215) At` to decay to `1//16^(th)` of its initially value isA. `400mu sec`B. `6.3 mu sec`C. `40mu sec`D. `300mu sec` |
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Answer» Correct Answer - A `A = A_(0)((1)/(2))^(n) rArr (A)/(A_(0))= ((1)/(2))^(n)` `(1)/(16) = ((1)/(2))^(4) = ((1)/(2))^(n)` `n = 4` `t = nT_(1//2) = 4xx100 = 400mu sec` |
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| 54. |
A nucelus with atomic number `Z` and neutron number `N` undergoes two decay processes. The result is a nucleus with atomic number `Z-3` and neutron `N-1`. Which decay processes took place?A. Two `beta^(bar)` decaysB. Two `beta^(+)` decaysC. An `alpha` decay and a `beta^(bar)` decaysD. `An `alpha` decay and a `beta^(+)` decays |
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Answer» Correct Answer - d Two protons and two neutrons are lost in an `alpha`-decays, so `Z` and `N` each decreases by `2`. A `beta^(bar)` decay changes a proton to a neutron, so `Z` decreases by `1` and `N` increases by `1`. The net result is `Z` decreases by `3` and `N` decreases by `1`. `._(Z)^(A)Xoverset(alpha-decay)(rarr)._(Z-2)^(A-4)Yoverset(beta-decay)(rarr)._(Z-3)^(A-4)Z` Initially, number of neutrons `N_i=(A-Z)` Now, number of neutrons `N_(f)=A-4-Z+3=N_(i)-1`. |
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| 55. |
Atomic number of a nucleus is `Z` and atomic mass is `M`. The number of neutron is.A. `M - Z`B. MC. ZD. `M + Z` |
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Answer» Correct Answer - A (a) `N = M - Z =` Total no. Of nucleons - no. of protons. |
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| 56. |
The force acting between proton and proton inside the nucleus is.A. CoulombicB. NuclearC. BothD. None of these |
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Answer» Correct Answer - C ( c) Both Coulomb and nuclear force act inside the nucleus. |
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| 57. |
A deutron strikes `._8 O^16` nucleus with subsequent emission of an alpha particle. Idenify the nucleus so produced.A. `._3 Li^7`B. `._5 B^10`C. `._7 n^13`D. `._7 N^14` |
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Answer» Correct Answer - D (d) `._8 O^16 + ._1H^2 rarr ._2 He^4 + ._7N^14`. |
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| 58. |
The mass of a `a_3^7 Li` nucleus is `0.042 u` less than the sum of the masses of all its nucleons. The binding energy per nucleon of `._3^7 Li` nucleus is nearly.A. 46 MeVB. 5.6 MeVC. 3.9 MeVD. 23 MeV |
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Answer» Correct Answer - B (b) If `m = 1u, c = 3 xx 10^8 ms^-1`, then `E = 931 MeV` i.e., `1u = 931 MeV` Binding energy `xx 0.042 xx 931 = 39.10 MeV` ` :.` Binding energy per nucleon =`(39.10)/(7) = 5.58 ~~ 5.6 MeV`. |
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| 59. |
The half-life of a radioactive isotope `X` is `50` years. It decays to another element `Y` which is stable. The two elements `X` and `Y` were found to be in the ratio of `1 : 15` in a sample of a given rock. The age of the rock was estimated to beA. 200 yrB. 250 yrC. 100 yrD. 150 yr |
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Answer» Correct Answer - A (a) We know that `(N)/(N_0) = ((1)/(2))^(t//t_1//2) rArr (1)/(16) = ((1)/(2))^(t//50)` `t = 4 xx 50 = 200 yr`. |
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| 60. |
The mass of nucleus `._(z)X^(A)` is less than the sum of the masses of `(A-Z)` number of neutrons and `Z` number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass `M` can break into two light nuclei of mass `m_(1)` and `m_(2)` only if `(m_(1)+m_(2)) lt M`. Also two light nuclei of massws `m_(3)` and `m_(4)` can undergo complete fusion and form a heavy nucleus of mass `M` 'only if `(m_(3)+m_(4)) gt M`'. The masses of some neutral atoms are given in the table below. `|{:(._(1)^(1)H,1.007825u,._(1)^(2)H,2.014102u,),(._(1)^(3)H,3.016050u,._(2)^(4)H,4.002603u,),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,),(._(30)^(70)Zn,69.925325u,._(34)^(82)Se,81.916709u,),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.97445u,),(._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u,):}|` The kinetic energy ( in `KeV`) of the alpha particle, when the nucleus at rest undergo alpha decay, isA. `5319`B. `5422`C. `5707`D. `5818` |
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Answer» Correct Answer - A `._(84)^(210)Po rarr_(2)^(4)He+_(82)^(206)Pb` `Q=(209.982876-4.002603-205.97455)C^(2)` `=5.422 MeV` From conservation of momentum. `:. K_(1)=5.319 MeV=5319 KeV` |
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| 61. |
Which of the following is correct regarding Nuclear fusion? (i) When two lighter nuclei moving at very high speeds fuse together to from a single heavier nucleus, then this phenomenon is called ‘Nuclear Fusion’ (ii) Solar energy is due to fusion. at sun, major process of fusion is carbon-carbon cycle and proton-proton cycle. Hydeogen bomb is based on fusion (iii) For fusion take place, a very high temperature of the order of `10^(8)K` is needed (iv) Some example of fusion are `._(1)H^(2) +._(1)n^(2)rarr._(2)He^(3) +._(0)n^(1) + "energy"` `._(2)He^(3) +._(1)H^(2)rarr._(2)He^(4) +._(1)H^(1) +"energy"`A. `(i),(ii),(iii)`B. `(i),(ii),(iv)`C. `(ii),(iii),(iv)`D. all |
| Answer» Correct Answer - D | |
| 62. |
The coolant in the nuclear reactor isA. Liquid sodiumB. cadmiumC. DeuteriumD. Liquid hydrogen |
| Answer» Correct Answer - A | |
| 63. |
When the nucleus o fan eletrically neutral atom undergoes a radioactive decay process , it will remain after he decay if the process isA. An `alpha` decayB. A `beta^(-)` decayC. A `gamma` decayD. A K-capture process |
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Answer» Correct Answer - C,D |
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| 64. |
Two identical samples (same material and same amout) `P and Q` of a radioactive substance having mean life `T` are observed to have activities `A_P` and `A_Q` respectively at the time of observation. If `P` is older than `Q`, then the difference in their age isA. `T l n ((A_(p))/(A_(Q))`B. `T l n ((A_(Q))/(A_(P))`C. `I/T l n ((A_(P))/(A_(Q))`D. `T((A_(P))/(A_(Q))` |
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Answer» Correct Answer - B |
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| 65. |
Two identical samples (same material and same amout) `P and Q` of a radioactive substance having mean life `T` are observed to have activities `A_P` and `A_Q` respectively at the time of observation. If `P` is older than `Q`, then the difference in their age isA. `T l n((A_(p))/(A_(Q)))`B. `T l n((A_(Q))/(A_(p)))`C. `(1)/(T)((A_(p))/(A_(Q)))`D. `T((A_(p))/(A_(Q)))` |
| Answer» Correct Answer - B | |
| 66. |
In a fission reaction `._92^236 U rarr ^117 X + ^117Y + n + n`, the binding energy per nucleon of `X` and `Y` is `8.5 MeV` whereas of `.^236 U` is `7.6 MeV`. The total energy liberated will be about.A. 200 KeVB. 2 MeVC. 200 MeVD. 2000 MeV |
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Answer» Correct Answer - C ( c) `Delta E = 8.5 xx 234 - 7.6 xx 236 = 195 = 195.4 MeV` =`200 MeV`. |
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| 67. |
In a fission reaction `._92^236 U rarr ^117 X + ^117Y + n + n`, the binding energy per nucleon of `X` and `Y` is `8.5 MeV` whereas of `.^236 U` is `7.6 MeV`. The total energy liberated will be about.A. `212 eV`B. `212 MeV`C. `2.12 MeV`D. `0.9 MeV` |
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Answer» Correct Answer - B `Q=BE_("product")-BE_("reactant")` |
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| 68. |
The nuclear charge (`Ze`) is non uniformlly distribute with in a nucleus of radius r. The charge density `rho(r)` (charge per unit volume) is dependent only on the radial distance r form the centre of the nucleus s shown in figure. The electric field is only along the radial direction. The electric field at `r=R` isA. independent of aB. directly proportional to aC. directly proportional to `a^(2)`D. inversely proportional to a |
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Answer» Correct Answer - A `E=(1)/(4 pi in _(0))(Q)/(R^(2)),Q=` total charge `e=Ze` `rArrE=(Ze)/(4 pi inR^(2))` `rArr E` at `r=R` is independent of a |
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| 69. |
Energy released in fusion of `1 kg` of deuterium nuclei.A. `8xx10^(13) J`B. `3xx10^(27) J`C. `2xx10^(7) "kWH"`D. `8xx10^(23) "MeV"` |
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Answer» Correct Answer - A |
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| 70. |
Nuclei of a radioactive element `X` are being produced at a constant rate `K` and this element decays to a stable nucleus `Y` with a decay constant `lambda` and half-life `T_(1//3)`. At the time `t=0`, there are `N_(0) `nuclei of the element X. The number `N_(Y)` of nuclei of `Y` at time `t` is .A. `Kt-(K-lambda N_(0))/(lambda)e^(-lambda t)+K-lambda(N_(0))/(lambda)`B. `Kt-(K-lambda N_(0))/(lambda)e^(-lambda t)+K-lambda(N_(0))/(lambda)`C. `Kt+(K-lambda N_(0))/(lambda)e^(-lambda t)`D. `Kt+(K-lambda N_(0))/(lambda)e^(-lambda t)` |
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Answer» Correct Answer - b `(dN_X)/(dt)=K-lambda N_(X)` N_(X)=(1)/(lambda)[K-K-lambda N_(0))e^(-lambda t)]` `(dN_(Y))/(dt)=lambdaN_(X)` `N_(Y)=Kt+(K-lambda N_0/lambda)e^(-lambda t) -(K-lambda N_(0))/(lambda)`. |
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| 71. |
Nuclei of a radioactive element `X` are being produced at a constant rate `K` and this element decays to a stable nucleus `Y` with a decay constant `lambda` and half-life `T_(1//3)`. At the time `t=0`, there are `N_(0) `nuclei of the element X. The number `N_(Y)` of nuclei of `Y` at time `t` is .A. `qt-((q-lambdaN_(0))/(lambda))e^(-lambda t)+(q-lambdaN_(0))/(lambda)`B. `qt+((q-lambdaN_(0))/(lambda))e^(-lambda t)`C. `qt+((q-lambdaN_(0))/(lambda))e^(-lambda t)+(1-lambdaN)/(lambda)`D. `qt-((q-lambdaN_(0))/(lambda))e^(-lambda t)` |
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Answer» Correct Answer - C `(dN_(Y))/(dt)=lambdaN_(X)=lambda(1)/(lambda)[q-(q-lambdaN_(0))e^(-lambda t)]` `rArr N_(Y)=qt+((1-lambdaN_(0))/(lambda))e^(-lambda t)- (q-lambdaN_(0))/(lambda)` |
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| 72. |
Nuclei of a radioactive element `X` are being produced at a constant rate `K` and this element decays to a stable nucleus `Y` with a decay constant `lambda` and half-life `T_(1//3)`. At the time `t=0`, there are `N_(0) `nuclei of the element `X`. The number `N_(X)` of nuclei of `X` at time `t=T_(1//2)` is .A. `(q+lambdaN_(0))/(2 lambda)`B. `(2lambdaN_(0)-1)(q)/(lambda)`C. `[lambdaN_(0)+(q)/(2)](1)/(lambda)`D. `[lambdaN_(0)-(q)/(2)](1)/(lambda)` |
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Answer» Correct Answer - A `(dN_(X))/(dt)=q-lambdaN_(X)rArrN_(X)=(1)/(lambda)[q-(q-lambdaN_(0))e^(-lambda t)]` Also, |
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| 73. |
Nucleus `A` decays to `B` with decay constant `lambda_(1)` and `B` decays to `C` with decay constant `lambda_(2)`. Initially at `t=0` number of nuclei of `A` and `B` are `2N_(0)` and `N_(0)` respectively. At `t=t_(o)`, no. of nuclei of `B` is `(3N_(0))/(2)` and nuclei of `B` stop changing. Find `t_(0)`?A. `(1)/(lambda_(1))ln((2lambda_(1))/(3lambda_(2)))`B. `(1)/(lambda_(1))ln((8lambda_(1))/(3lambda_(2)))`C. `(1)/(lambda_(1))ln((7lambda_(1))/(3lambda_(2)))`D. `(1)/(lambda_(1))ln((4lambda_(1))/(3lambda_(2)))` |
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Answer» Correct Answer - D `(dN_(B))/(dt)=lambda_(1)N_(A)-lambda_(2)N_(B)` `0= lambda_(1)(2N_(0))e^(-lambda t_(0))-lambda_(2)((3N_(0))/(2)), t_(0)=(1)/(lambda_(1)) ln((4 lambda_(1))/(3 lambda_(2)))` |
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| 74. |
One gram ofo a radioactive substance disintegrates at the rate of `3.7 xx10^(10)` disintegarations per second. The atomic mass of the subsatnce is `226`. Calculate its mean life. |
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Answer» Correct Answer - `7.194 xx10^(10)s` Given that activity of substance is `A_(c)=3.7 xx10^(10)dps` The number of atoms in `1 g` of substance, `N=(1 xx 6.023 xx10^(23))/(226) =2.66 xx10^(21)"atoms"` If `lambda` is the decay constant of the substance, we know that activity is given by `A_(C) =lambda N` or `lambda =(A_(c))/(N)` `=(3.7 xx10^(10))/(2.66 xx10^(21)) s^(-1)=1.39 xx 10^(-11) s^(-1)` Thus, mean life of the radioactive substance is `T_(m) =(1)/(lambda)=(1)/(1.39 xx10^(-11)) =7.194 xx 10^(10)s`. |
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| 75. |
The three stable isotopes of neon `._10Ne^(20), ._10Ne^(21)` and `._10Ne^(22)` have respective abundances of `90.51%,0.27%` and `9.22%`. The atomic masses of the three isotopes are `19.99u, 20.99u` and `21.99u` respectively. Obtain the average atomic mass of neon. |
| Answer» `m(90.51xx19.99+0.27xx20.99+9.22xx22)/(100)=20.18 u` | |
| 76. |
A radioactive source has a half life of 3 hours. A freshly prepared sample of the same emits radiation 16 times the permissible safe value. The minimum time after which it would be possible to work safely with the source is :A. 6 hoursB. 12 hoursC. 18 hoursD. 24 hours |
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Answer» Correct Answer - B |
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| 77. |
A radioactive element emits `200` particle per second. After three hours `25` particle per second are emitted. The half-life period of element will beA. 50 minutesB. 60 minutesC. 70 minutesD. 80 minutes |
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Answer» Correct Answer - B (b) `R = (dN)/(dt) prop N rArr (R_2)/(R_1) = (N_2)/(N_1)` But `(N_2)/(N_1) = ((1)/(2))^(t_1//2) rArr (25)/(200) = (1)/(8) = ((1)/(2))^3 rArr (t)/(t_(1//2)) = 3` `:. t_(1//2) = (t)/(3) = (3)/(3) = 1 hour = 60` minutes. |
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| 78. |
Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of the resulting radiation?A. `(h)/(2m_(0)C)`B. `(2h)/(m_(0)C)`C. `(l n 2)/(3t)`D. `(3 l n2)/(t)` |
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Answer» Correct Answer - C `(hc)/(lambda)+(hc)/(lambda)=m_(0)C^(2)+m_(0)C^(2)` |
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| 79. |
The miniumum amount of energy released in annihilation of electron-Positron isA. `1.02 MeV`B. `0.58 MeV`C. `185 MeV`D. `200 MeV` |
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Answer» Correct Answer - A `0.51MeV+0.51MeV=1.02MeV` |
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| 80. |
What would be the energy required to dissociate completely `1 g` of `Ca-40` into its constituent, particles? Given: Mass of proton `=1.00866 am u`, Mass of neutron `=1.00866 am u`, Mass of `Ca-40 =39.97454 am u`, (Take `1 am u =931 MeV`).A. ` 4.813 xx 10^(24)MeV`B. ` 4.813 xx 10^(24)eV`C. ` 4.813 xx 10^(23)MeV`D. none of these |
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Answer» Correct Answer - a Mass defect, `Deltam=20(1.007277 +1.00866)-39.97545` `=40.31874-39.97545` `=0.34329 a.m.u.` `:.` Binding energy `=0.34329 xx 931 =319 .6 MeV` When one atom of `Ca-40` completely dissociates, the energy to be supplied `=319.6 MeV` `1 g` of `Ca-40` contains `(6.023 xx10^(23))/(40)=1.506 xx 10^(22)` atoms The energy required for the dissocoation of `1 g` of `Ca-40` `319.6 xx 1.506 xx10^(22)` `4.813 xx 10^(24)MeV`. |
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| 81. |
The probability of a radioactive atoms to survive 5 times longer than its half-value period is-A. `2/5`B. 2x5C. `2^(-5)`D. `2^(5)` |
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Answer» Correct Answer - C `(N_(0)-N)/(N_(0)) =e^(-lambdat)` |
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| 82. |
After emission of an `alpha-`particle by a radiative element `._(84)X^(212)`, the resulting element would be-A. `._(84)Y^(210)`B. `._(82)Y^(208)`C. `._(84)Y^(208)`D. `._(86)Y^(210)` |
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Answer» Correct Answer - B `._(54)X^(212) to ._(52)Y^(208) +._(2)He^(4)` |
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| 83. |
The binding energies per nucleon for a deuteron and an `alpha-`particle are `x_1` and `x_2` respectively. What will be the energy `Q` released in the following reaction ? `._1H^2 + ._1H^2 rarr ._2He^4 + Q`.A. `4(x_1 + x_2)`B. `4(x_2 - x_1)`C. `2(x_1 + x_2)`D. `2(x_2 - x_1)` |
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Answer» Correct Answer - B (b) `Q = 4 (x_2 - x_1)`. |
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| 84. |
The binding energies per nucleon for a deuteron and an `alpha-`particle are `x_1` and `x_2` respectively. What will be the energy `Q` released in the following reaction ? `._1H^2 + ._1H^2 rarr ._2He^4 + Q`.A. `4(x_(1)+x_(2))`B. `4(x_(2)-x_(1))`C. `2(x_(1)+x_(2))`D. `2(x_(2)-x_(1))` |
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Answer» Correct Answer - B `._(1)H^(2) +._(1)H^(2) rarr._(2)He^(4) +Q` `Q -` value `= 4x_(2) - 4x_(1) = 4(x_(2)-x_(1))` |
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| 85. |
The binding energies per nucleon for a deuteron and an `alpha-`particle are `x_1` and `x_2` respectively. What will be the energy `Q` released in the following reaction ? `._1H^2 + ._1H^2 rarr ._2He^4 + Q`.A. `(x_(1)+x_(2))`B. `(x_(2)-x_(1))`C. `4(x_(1)-x_(2))`D. `4(x_(2)-x_(1))` |
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Answer» Correct Answer - D `(4x1.0089+5x1.009)-9.012=Deltam` binding energy `=Deltamxxc^(2)` Bindng energy /nucleon `=("Binding energy")//9=6.72 MeV` |
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| 86. |
If the mass of proton= 1.008 a.m.u. and mass of neutron=1.009a.m.u. then binding energy per nucleon for `._(4)Be^9` (mass=9.012 amu) would be-A. 0.065 MeVB. 60 .44MeVC. 67.2 MeVD. 6.72 MeV |
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Answer» Correct Answer - B Rest mass energy+kinetic energy =energy of gama photon `0.51xx2MeV+2xxE_(k)=2.42 MeV`. `E_(k)=0.7 MeV` |
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| 87. |
A helium atom, a hydrogen atom and a neutron have mases of` 4.003 u, 1.008 u` and `1.009 u` (unified atomic mass units), respectively. Assuming that hydogen atoms and neutrons can be fuse to from helium, what is the binding energy of a helium nucleus?A. `2.01 u`B. `3.031 u`C. `1.017 u`D. `0.031u` |
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Answer» Correct Answer - d Refer to the defination of mass defect. |
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| 88. |
A radioactive sample has `8.0xx10^(18)` active nuclei at a certain instant. How many of these nuclei will still be in the active state after two half-life `("in" xx10^(18))`? |
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Answer» Correct Answer - 2 In one half-life the number of active nuclei reduces to half the original number. Thus, in two half-lives the numbers is reduced to `((1)/(2))((1)/(2))` of the original number. The number of remaining active nuclei is, therefore, `8.0xx10^(18)xx((1)/(2))=2xx10^(18)`. |
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| 89. |
A radioactive sample decays with an average life of `20 ms`. A capacitor of capacitance `100 muF` is charged to some potential and then the plates are connected through a resistance `R`. What should be the value of `R` so that the ratio of the charge on the capacitor to the activity of the radioactive sample remains constant in time? |
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Answer» The activity of the sample at time `t` is given by `A=A_(0)e^(-lambdat)` where `lambda` is the decay constant and `A_(0)` is the activity at time `t=0` when the capacitor plates are connected. The charge on the capacitor at time `t` is given by where `Q_(0)` is the charge at `t=0` and `C=100 muF` is the capacitance. Thus, `(Q)/(A)=(Q_(0))/(A_(0))(e^(-t//CR))/(e^(-lambdat))` It is independent of `t` if `lambda=(1)/(CR)` or `R=(1)/(lambda C)=(t_(av))/(C )=(20xx10^(-3)s)/(100xx10^(-6)F)=200 Omega` |
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| 90. |
A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life `tau`.Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.A. `(tau)/(C )`B. `(2tau)/(C )`C. `(tau)/(2C)`D. `(3tau)/(2C)` |
| Answer» Correct Answer - B | |
| 91. |
A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life `tau`.Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.A. `(2t)/( C )`B. `(C )/(2t)`C. `2t C`D. `t C` |
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Answer» Correct Answer - A `(U_(c ))/(A)="constant" , ((1)/(2)CV_(0)^(2))/(lambdaN_(0))=((1)/(2)CV^(2))/(lambda N)` Where `V-V_(0)^(-t//CR), N=N_(o)e^(-lambda t)` and `lambda=(1)/(t)` |
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| 92. |
A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life `tau`.Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time. |
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Answer» `bar(T) = tau` `lambda = (1)/(bar(T)) = (1)/(tau)` Activity of radioactive smaple at time `t` `A = A_(0)e^(-lambdat) = A_(0)e^(-t//tau) ….(i)` In discharging `RC` circuit `q = q_(0)e^(-t//RC)` Energy stored in capacitor `U = (q^(2))/(2C) = (q_(0)^(2))/(2C)e^(-(2t)/(RC))` `(U)/(A) =((q^(2))/(2C)e^(-(2t)/(RC)))/(A_(0)e^(-t//tau))` If `U//A` is independent of time, `e^(-2t//RC) = e^(-t//tau)` `(2)/(RC) = (1)/(tau)` `R =(2tau)/(C )` |
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| 93. |
A radioactive sample has `6.0 xx 10^18` active nuclei at a certain instant. How many of these nuclei will still be in the same active state after two half-lives? |
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Answer» Correct Answer - `[1.5 xx 10^(18)]` |
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| 94. |
(a) A radioactive sample has `6.0 xx 10^(18)` active nuclei at a certain instant. How many of these nuclei will still be the same active state after two half-lives? (b) Radioactive phosphours`-32` has a half-life of `14` days. A source containing this isotope has an initial activity of `10muCi`. (i) What is the activity of the source after `42` days? (ii) What time elapse before the activity of the source falls at `2.5muCi`? (c) `1g` of a radioactive substance disintergrates at the rate of `3.7 xx 10^(10) dps`. The atomis mass of the substance is `226`. Calculate its mean life. |
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Answer» (a) `N = N_(0)((1)/(2))^(n)` `n = 2` `N = (N_(0))/(4) = (6xx10^(18))/(4) = 15 xx 10^(17)` (b) `t = nT_(1//2)` `42 = nxx14 rArr n = 3` (i) `A = A_(0) ((1)/(2))^(n) = A_(0)((1)/(2))^(3) = (A_(0))/(8) = (10)/(8) muCi = 1.25muCi` (ii) `A = A_(0) ((1)/(2))^(n)` `2.5 = 10((1)/(2))^(n) rArr (1)/(4) = ((1)/(2))^(n)` `n =2` `t = nT_(1//2) = 2xx14 = 28` days (c) `N = (m)/(M) N_(A)` `A = lambdaN = (1)/(barT)(mN_(A))/(M)` `bar(T) = (mN_(A))/(MA) = (1xx6.023xx10^(23))/(226xx3.7xx10^(10)) sec` =` 7.2 xx 10^(10) sec` |
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| 95. |
Calculate the binding energy of an alpha particle from the following data: `mass of _1^1H atom = 1.007825 u` mass of neutron `= 1.008665 u` mass of `_4^2He atom = 4.00260 u`. Take `1 u = 931 MeV c^(-2)` |
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Answer» The alpha particle contains`2` protons and `2` neutrons. The binding energy is `B =(2 xx 1.007826 u + 2 xx 1.008665 u -4.00260 u)c^(2)` `=(0.03038 u)c^(2)` `=0.03038 xx 931 MeV =28.3 MeV`. |
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| 96. |
The radius of the nuclues is proprtional to, (if `A` is the atomic mass number)A. `A`B. `A^(3)`C. `A^(1//3)`D. `A^(2//3)` |
| Answer» Correct Answer - C | |
| 97. |
Calculte the binding energy per nucleon for `._(10)^(20)Ne,._(26)^(56)Fe` and `._(98)^(238)U`. Given that mass of neutron is `1.008665 am u`, mass of proton is `1.007825 am u`, mass of `._(10)^(20)Ne` is `19.9924 am u`, mass of `._(26)^(56)Fe` is `55.93492 am u` and mass of `._(92)^(238)U` is `238.050783 am u`. |
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Answer» Correct Answer - `7.57 MeV` Binding energy of nucleus is given by the equation `B(._Z^AX) =[(A -Z)m_n +Z m_(p) - M(._Z^AX)]c^(2)` On dividing binding energy by the mass number, we obtain the binding energy per nucleon. `B(._(10)^(20)Ne) =[10 m_(n) +10 m_(p) - M(._(10)^(20) Ne)]c^(2)` `=[10 xx 1.008665 +10 xx 1.007825 -55.93492]` `xx 931.5` `=492 MeV` Hence, binding energy per nucleon `=(B(._(26)^(56)Fe))/(56 )=8.79` `MeV//nuclon` Binding energy for `._(92)^(238)U` is `[146m_(n) +92m_(p) -M(._(92)^(238)U)]c^(2)` `=[146 xx 1.008665 +92 xx 1.007825 -238.050783]` `xx9315` `=1802 MeV` Binding enregy per nucleon `=(B(._(92)^(238)Fe))/(238)=(1802)/(238) =7.57 MeV`. |
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| 98. |
A town has a population of 1 million. The average electric power needed per person is 300W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%.(a) Assuming 200 MeV of thermal energy to come form each fission event on an average, find the number of events on an place every day. (b) Assuming the fission to take place largely through `U^235`, at what rate will the amount of `U^235` decrease ? Express uour answer in kg per day. (c) Assuming that uranium enriched to 3% in `U^235` will be used, how much uranium is needed per month (30 days)? |
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Answer» Power required `= 10^(6) xx 300 = 3 xx 10^(8)W` Input power, `P = (3xx10^(8))/(0.25) = 12xx10^(8)W` (a) Let number of events/day `= n` `(nxx200xx1.6xx10^(-13))/(24xx3600) = 12 xx 10^(8)` `n = 3.24 xx 10^(24)` (b) `n = (m)/(M) N_(A)` `3.24xx10^(24) = (m)/(235) xx 6.023 xx 10^(23)` `m = 126.4 xx 10 = 1264g//day` `=1.264kg//day` (c) Total `U^(235)` used in `30` days `= 1.264 xx 30kg` Uranium enriched `3%` Uranium needed `= 1.264 xx 30 xx (100)/(3)` `= 1264kg` |
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| 99. |
When the number of nucleons in a nuclues increases the binding energy per nucleonA. Increase continuously with mass numberB. Decreases continuosly with mass numberC. Remains constant with mass numberD. First increases and then decreases with increase in mass number |
| Answer» Correct Answer - D | |
| 100. |
If mass of `U^(235)=235.12142 a.m.u.`, mass of `U^(236) =236.1205 amu`, and mass of neutron `=1.008665 am u`, then the energy required to remove one neutron from the nucleus of `U^(236)` is nearly about.A. `75 MeV`B. `6.5 MeV`C. `1 eV`D. zero |
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Answer» Correct Answer - b Mass of one atom of `U.^(235)` is `235.121420 a.m.u.` Mass of one neutron `=1.008665 a.m.u.` Sum of the masses of `U.^(235)` and neutron `=236. 130085 =236.130 a.m.u.` Mass of one of `U.^(236)` is `236.123050 a.m.u.` `=236.123 a.m.u.` Mass defect `=236.136 -236.123` `=0.007a.m.u.` Therefore, enregy required to remove one neutron is `0.007 xx 931 MeV=6.517 MeV=6.5MeV`. |
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