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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.
| 201. |
IN a nuclear reactor, fission is produced in 1 g of `.^(235)U` `(235.0349 am u)`. In assuming that `._(53)^(92)Kr (91.8673 am u)` and `._(36)^(141)Ba (140.9139 am u)` are produced in all reactions and no energy is lost, calculate the total energy produced in killowatt. Given: `1 am u =931 MeV`. |
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Answer» Correct Answer - `2.28 xx 10^(4)kWh` The nuclear fission reaction in the above process is `._(92)^(235)U +._(0) + ._(0)^(1)n rarr._(56)^(141)Ba +._(36)^(92)Kr+3 (._(0)^(1)n` Thesum of the mass before reaction is `235.0439 +1.007=236.0526 a.m. u.` The sum of the masses after reaction is `140.9139 +91.8973 +(1.0087) =235.8375 a.m.u.` Mass defect, `Delta m =236.0526 -235. 8373 =0.2175 a.m.n` Energy released in the fission of `.^(235)U` nulceus is given by `DeltaE=Deltam xx 931.2 MeV` `=0.2153 xx 931.2 =200 MeV` Number of atoms in `1 g` of `.^(235)U` `N=(6.02 xx10^(23))/(235)=2.56 xx10^(21)` Energy released in fission of `1g` of `.^(235)U`, `E=200 xx 2.56 xx 10^(21) MeV` `=5.12 xx10^(23) MeV =(5.12 xx 10^(23)) xx (1.6 xx10^(-13))` `=8.2 xx 10^(10)J` `(8.2 xx 10^(10))/(3.6 xx 10^(6) kWh =2.28 xx10^(4) kWh` |
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| 202. |
The half life period of `Pb^(210)` is 22 years. If `2 g` of `Pb^(210)` is taken, then after 11 years the amount of `Pb^(210)` will be present isA. `0.1414 g`B. `1.414 g`C. `2.828 g`D. `0.707 g` |
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Answer» Correct Answer - B `t=(t_((1)/(2)))/(log 2)((N_(0))/(N))` |
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| 203. |
In an experiment on two radioactive isotopes of an element (which do not decay into each other), their number of atoms ratio at a given instant was found to be `3`. The rapidly decaying isotope has large mass and an activity of `1.0muCi` initially. The half-lives of the two isotopes are known to be `12` hours and `16` hours. what would be the activity of each isotope and their number of atoms ration after two days? |
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Answer» `((N_(A))/(N_(B)))_(0) =3` `(A_(A))_(0) = 1muCi` `(T_(1//2)) = 12h, (T_(1//2))B = 16` `((A_(A))_(0))/((A_(B))_(0))=(lambda_(A)(N_(A))_(0))/(lambda_(B)(N_(B))_(0))=((T_(1//2))_(B))/((T_(1//2))_(A))((N_(A))/(N_(B)))_(0)` `(1)/(A_(B))_(0) = (16)/(12) xx 3 rArr (A_(B))_(0) = (1)/(4)muCi` `t = 2 days = 2 xx 24 = 48h` where `A:` number of half lives in `2` days `= 48//12 = 4`. `A_(A) = (A_(A))_(0)((1)/(2))^(n) =1((1)/(2))^(4) = (1)/(16)muCi` `B:` number of half lives in `2` days `= 48//16 = 3`. `A_(B) = (A_(B))_(0)((1)/(2))^(n) =(1)/(4)((1)/(2))^(3) = (1)/(32)muCi` `(A_(A))/(A_(B)) = ((T_(1//2))_(B))/((T_(1//2))_(A))(N_(A))/(N_(B))` `(1//16)/(1//32) = (16)/(12)(N_(A))/(N_(B)) = (4)/(3)(N_(A))/(N_(B))` `(N_(A))/(N_(B)) = (3)/(2)` Alternatively `N_(A) = (N_(A))_(0)((1)/(2))^(n_(1)) = (N_(A))_(0)((1)/(2))^((48)/(12)) = (N_(A))_(0)((1)/(2))^(4)` `N_(B) = (N_(B))_(0)((1)/(2))^(n_(2)) = (N_(B))_(0)((1)/(2))^((48)/(16)) = (N_(B))_(0)((1)/(2))^(3)` `(N_(A))/(N_(B)) = ((N_(A))/(N_(B)))_(0) ((1)/(2)) = (3)/(2)` |
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| 204. |
A radioactive nucleus with `Z` protons and `N` neutrons emits an `alpha-`particle, `2 beta-`particle and `2 gamma` rays. The number of protons and neutrons in the nucleus left after the decay respectively, areA. `Z - 3,N - 1`B. `Z - 2, N - 2`C. `Z - 1, N - 3`D. `Z,N - 4` |
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Answer» Correct Answer - D (d) With emission of an `alpha` particle `(._2He^4)` mass number decreases by `4` unit and atomic number decrease by `2` units and with emission of `2 beta^-1` particle atomic number increases by `2` units. So `Z` will remain same and `N` become `N-4`. |
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| 205. |
In which of the following decays the atomic number decreases? (i) `alpha`-decay (ii) `beta^(+)`-decay (iii) `beta^(-)`-decay (iv) electron captureA. `(i),(ii),(iii)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iv)`D. all |
| Answer» Correct Answer - C | |
| 206. |
A free neutron decays to a proton but a free proton does not decay to a neutron. This is becauseA. neutron is a composite particle made of a proton and an electron whereas proton is fundamental particleB. neutron is an uncharged particle whereas proton is a charge particleC. neutron has large mass than protonD. weak forces can operate in a neutron but not in a proton |
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Answer» Correct Answer - C As a proton is lighter than a neutron, proton can not be converted into neutron without providing energy from outside. Reverse is possible. The weak interaction force is responsible in both the process (i) conversion of `p` to `n` and (ii) conversion of `n` to p p |
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| 207. |
The electron emitted in beta radiation originates fromA. Inner orbits of atomsB. Free electrons existing in nucleiC. Decay of neutron in a nucleusD. Photn escaping from the nucleus |
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Answer» Correct Answer - C |
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| 208. |
The electron emitted in beta radiation originates fromA. Inner orbits of atomsB. Free electrons existing in nucleiC. Decay of a neutron in a nucleusD. Photon escaping from the nucleus |
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Answer» Correct Answer - C ( c) During `beta-`decay, a neutron is transformed into a proton and an electron. |
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| 209. |
During a negative beta decay,A. an atomic electorn is ejectedB. an electron which is already present within the nucleus is ejectedC. a neutron in the nucleus decays emitting an electornD. a proton in the nucleus in beta decays emitting an electorn |
| Answer» Correct Answer - C | |
| 210. |
Proton and singly ionized of `U^(235)` & `U^(238)` are passed in turn (which means one after the other and not at the same time) through a velocity selectro and then enter a uniform magnetic field. The protons describe semicircles of radius `10 mm`. The separation between the ions of `U^(235)` and `U^(238)` after describing semicircle is given by A. `60 mm`B. `30 mm`C. `2350 mm`D. `2380 mm` |
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Answer» Correct Answer - A `R=(mv)/(qB)` `R_(P)=(m_(P)v)/(eB)` `R_(235_(upsilon))=(m_(235_(upsilon).)v)/(eB)` `Deltax=2xx(3M_(P)v)/(eB)=6xx10mm=60mm`. |
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| 211. |
The electron emitted in beta radiation originates fromA. inner orbits of atomsB. free electorns exiting in nucleiC. decay of a neutron in a nucleiD. photon escaping form the nucleus |
| Answer» Correct Answer - C | |
| 212. |
Consider a sample of a pure beta-active materialA. All the beta particles emitted have the sae energyB. The beta particles originally exist inside the nucleus and are ejected at the time of beta decayC. The antineutrion emitted in a beta decay has zero mass and hence zero momentumD. The active nucleus changes to one of its isobars after the beta decay |
| Answer» Correct Answer - D | |
| 213. |
In an `alpha` -decay, the kinetic energy of `alpha`-particles is `48 MeV` and `Q` value of the reaction is `50 MeV`. The mass number of the mother nucleus is (assume that daughter nucleus is in ground state) |
| Answer» `KE_(alpha)=((A-4)/(A))|Q|rArrA=100` | |
| 214. |
When a `beta^(-)` particle is emitted from a nucleus, the neutrons-proton ratio:A. is decreasedB. is increasedC. remains the sameD. first (A) then (B) |
| Answer» Correct Answer - A | |
| 215. |
Consider a sample of a pure beta-active materialA. All the beta particles emitted have the same energyB. The beta particles originally exist inside the nucleus and are ejected at the time of the beta decayC. The antineutrino emitted in a beta decay has zero rest mass and hence zero momentum.D. The active nucleus changes to one of its isobars after the beta decay |
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Answer» Correct Answer - D (A) The emitted `beta`-particles may have varying energy. (B) `e^(-)` or `e^(+)` does not exists inside the nucleus. (C ) `overset(-)v` does carry momentum. (D) In `beta`-decay mass number does not change. |
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| 216. |
In an `alpha` -decay, the kinetic energy of `alpha`-particles is `48 MeV` and `Q` value of the reaction is `50 MeV`. The mass number of the mother nucleus is (assume that daughter nucleus is in ground state)A. `96`B. `100`C. `104`D. none of these |
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Answer» Correct Answer - B We have `K_(alpha)=(m_(y))/(m_(y)+m_(alpha)).Q " "rArr K_(alpha)=(A-4)/(A).Q" "rArr48=(A-4)/(A).50 " "rArrA=100` |
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| 217. |
Consider a particle , `beta` particle and `gamma - rays` , each having an energy of `0.5 MeV` . In increase order of panetrating power , the radiation are .A. `alpha,beta,gamma`B. `alpha,gamma ,beta`C. `beta,gamma,alpha`D. `gamma,beta,alpha` |
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Answer» Correct Answer - c The penetrating power is dependent on velocity. For a given energy, the velocity of `gamma`-radiation is highest and `alpha`-particle is least. |
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| 218. |
Masses of two isobars `._(29)Cu^(64)` and `._(30)Zn^(64)` are `63.9298 u` and `63.9292 u`, respectively. It can be concluded from these data that .A. Both the isobars are stableB. `.^(64)Zn` is radioactive, decaying to `.^(64)Cu` through `beta`-decayC. `.^(64)Cu` is radioactive, decaying to `.^(64)Zn` through `gamma`-decayD. `.^(64)Cu` is radioactive, decaying to `.^(64)Zn` through `beta`-decay |
| Answer» Correct Answer - D | |
| 219. |
Masses of two isobars `._(29)Cu^(64)` and `._(30)Zn^(64)` are `63.9298 u` and `63.9292 u`, respectively. It can be concluded from these data that .A. both the isobars are stableB. `Zn^(64)` is radioactive , decaying to `Cu^(64)` through `beta`-decayC. `Cu^(64)`is radioactive , decaying to `Zn^(64)` through `gamma`-decayD. `Cu^(64)`is radioactive , decaying to `Zn^(64)` through `beta`-decay |
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Answer» Correct Answer - d By the conservation of charge and nucleons, only potential is feasible. |
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| 220. |
Half life of a radio active element is `5 min`. `10 sec`. Time taken for `90%` of it to disintergrate is nearlyA. `100 min`B. `1000 sec`C. `10^(4) sec`D. `10^(4) min` |
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Answer» Correct Answer - B `t=(t_(1/(2)))/(log 2) log((N_(0))/(N))` |
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| 221. |
The half-life period of a radio-active element `X` is same as the mean life time of another radio-active element `Y`. Initially they have the same number of atoms. Then:A. X and Y have the same decay rate initially.B. X and Y decay at the same rate alwaysC. Y will decay at a faster rate than XD. X will decay at a faster rate than y |
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Answer» Correct Answer - D |
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| 222. |
The half-life period of a radio-active element `X` is same as the mean life time of another radio-active element `Y`. Initially they have the same number of atoms. Then:A. `X` will decay faster than `Y`B. `Y` will decay faster then `X`C. `X` and `Y` have same decay rate initiallyD. `X` and `Y` decay at same rate always |
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Answer» Correct Answer - B `( l n2)/(lambda_(x))=(1)/(lambda_(gamma))=lambda_(gamma)=1.4lambda_(x), lambda_(gamma)gtlambda_(X), Y` will decay fater than `X` |
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| 223. |
This question contains Statement - 1 and Statement -2 Of the four choice given after the Statements , choose the one that best decribes the two Statements Statement- 1: Energy is released when heavy undergo fission or light nuclei undergo fusion and Statement- 2: for nuclei , Binding energy nucleon increases with increasing `Z` while for light nuclei it decreases with increasing `Z`A. Statement-1 is true, Statement-2 is a correct explanantion for statement -1B. Statement -1 is true, Statement-2 is true, Statement-2 is not a correct explanation for statement -1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
| Answer» Correct Answer - C | |
| 224. |
The `alpha`-particle areA. high energy electronsB. positively charged hydrogen ionsC. high energy `alpha`-radiationD. doubly positively charged helium nuclei |
| Answer» Correct Answer - D | |
| 225. |
`alpha`-particles carriesA. Mass `1`B. Mass `2`C. Mass `3`D. Mass `4` |
| Answer» Correct Answer - D | |
| 226. |
A `beta-` particle is emitted by a radioactive nucleus at the time of convertsion of a:A. Neutron into a protonB. Proton into a neutronC. Nucleon into energyD. Positron into energy |
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Answer» Correct Answer - A |
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| 227. |
Half life of a radio-active substance is `20` minutes. The time between `20 %` and `80 %` decay will beA. 20 minutesB. 40 minutesC. 30 minutesD. 25 minutes |
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Answer» Correct Answer - B |
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| 228. |
The decay constant of a radio active element, which disintergrates to `10 gms` from `20 gms` in `10` minutes isA. `0.693_("min"^(-1))`B. `6.93_("min"^(-1))`C. `0.693_("sec"^(-1))`D. `0.0693_("min"^(-1))` |
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Answer» Correct Answer - D `lambda=0.693//T_((1)/(2))` |
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| 229. |
A radio active isotope having a half life of `3` days was received after `9` days. It was found that there was only `4 gms` of the isotope in the container. The initial weight of the isotope when packed wasA. `8 g`B. `64 g`C. `48 g`D. `32 g` |
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Answer» Correct Answer - D `n=(t)/(t_(1/2))=(9)/(3)= 3, W=(W_(0))/(2^(n))` `W_(0)=2^(n),W=2^(3)(4)=32 gm` |
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| 230. |
The half-life period of radium is `1600` years. The fraction of a sample of radium that would remain after `6400` years is.A. `(1)/(4)`B. `(1)/(2)`C. `(1)/(8)`D. `(1)/(16)` |
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Answer» Correct Answer - D (d) Fraction `= (N)/(N_0) = ((1)/(2))^((6400)/(1600)) = ((1)/(2))^4 = (1)/(16)`. |
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| 231. |
The half life of a radio active substance is `6` hours. The amount of the substance undergone disintegration when `36 gms` of it undergoes decay for `18` hours isA. `31.5 gm`B. `4.5 gm`C. `18 gm`D. `9 gm` |
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Answer» Correct Answer - A `W=(W_(0))/(2^(n)), n=3` |
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| 232. |
The half-life period of radium is `1600` years. The fraction of a sample of radium that would remain after `6400` years is.A. `1/4`B. `1/2`C. `1/8`D. `1/16` |
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Answer» Correct Answer - D |
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| 233. |
Find the energy of the reaction `.^(14)N(alpha, p) .^(17)O`, if the kinetic energy of the incoming `alpha`-particle is `T_(alpha) = 4.0 MeV` & the proton outgoing at an angle `theta = 60^(@)` to the motion direction of the `alpha`-particle has a kinetic energy `T_(p) = 2.09 MeV`. |
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Answer» Correct Answer - `[Q = (1-eta_(p))T_(p) - (1-eta_(alpha))T_(alpha)-2 sqrt(eta_(p)eta_(alpha)T_(p)T_(alpha)) cos theta =-12 MeV, "where" n_(p) = m_(p)//m_(o), n_(alpha) = (m_(alpha))/(m_(0))]` |
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| 234. |
(a) The half life period of radium is 1590 yrs. After how many years will one gram of the pure element, (i)be reduced to one centigram, (ii) lose one centigram. (b) The half life of radon is 3.8 days. After how many days will only one twentieth of radon sample be left over? (c) 1 gm of radioactive substance takes 50 sec. to lose 1 centigram. Find its half life period? |
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Answer» Correct Answer - [(a) 10560 yrs, 23.25 yrs. (b) 16.45 days (c) 57.75 min.] |
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| 235. |
Half life period of radium is `1600` years. `2 gm` of radium undergoes decay and gets reduced to `0.125 gms` inA. `3200` yearsB. `25600` yearsC. `8000` yearsD. `6400` years |
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Answer» Correct Answer - D `N=N_(0)e^(-lambda t)` |
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| 236. |
An alpha particles kinetic energy `T= 5.3 MeV` initiates a nuclear reaction `Be^(9)(alpha,n)C^(12)` with energy yield `Q=+5.7MeV`. Find the kinetic energy of the nerutron outgoing at right angle to the motion direction of the alpha-particles. |
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Answer» Correct Answer - `[T_(n) = (Q + (1 -m_(a)//m_(c))T)/(1 + m_(n)//m_(c)) = 8.5 MeV]` |
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| 237. |
Find the quantity of polonium `._(80)^(210)Po` whose activity is `3.7xx10^(10)` dps. Find also the number of atoms of polonium disintegrated during its mean life. (Half life of polonium is 138 days) |
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Answer» Correct Answer - `[0.22 mg, 4.03xx10^(17)]` |
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| 238. |
There are two radio nuceli `A` and `B. A` is an `alpha` emitter and `B` a `beta` emitter. Their disintegration constant are in the ratio of `1:2` What should be the ratio of number of atoms of `A` and `B` at any time t so that probabilities of getting alpha and beta particles are same at that instant?A. `2:1`B. `1:2`C. `e`D. `re^_(1)` |
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Answer» Correct Answer - a `lambda_(A) lambda_(B) =(1)/(2)` Probabilites of getting `alpha`- and `beta`- particles are same. Thus, rates of disintegration are equal. `:. lambda_(A) N_(A) =lambda_(B) N_(B)` or `(N_(A))/(N_(B)) = (lambda_(B))/(lambda_(A))=2` . |
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| 239. |
The energy of alpha particles emitted by `.^(210)Po` is 5.3 MeV. The half life of this alpha emitter is 138 days. (a) What mass of `.^(210)Po` is needed to power a thermoelectric cell of 1 W ouput if the efficiency of energy conservation is 8 percent? (b) What would be the power output after 1 year ? |
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Answer» Correct Answer - [88.4 mg, 0.16 watt] |
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| 240. |
A radioactive sample is `alpha-`emitter with half life `138.6` days is observed by a student to have `2000` disintegration/sec. The number of radioactive nuclei for given activity are.A. `3.45 xx 10^10`B. `1 xx 10^10`C. `3.45 xx 10^15`D. `2.75 xx 10^11` |
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Answer» Correct Answer - A (a) Activity of substance that has `2000` disintegration/sec `=(2000)/(3.7 xx 10^10) = 0.054 xx 10^-6 ci = 0.054 mu ci` The number of radioactive nuclei having activity `A` `N = (A)/(lamda) = (2000 xx T_(1//2))/(log_e 2)` =`(2000 xx 138.6 xx 24 xx 3600)/(0.693) = 3.45 xx 10^10`. |
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| 241. |
The life-life of `Bi^210` is `5` days. What time is taken by `(7//8)^th` part of the sample of decay ?A. 3.4 daysB. 10 daysC. 15 daysD. 20 days |
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Answer» Correct Answer - C |
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| 242. |
The life-life of `Bi^210` is `5` days. What time is taken by `(7//8)^th` part of the sample of decay ?A. `3.4 daysB. 10 daysC. 15 daysD. 20 days |
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Answer» Correct Answer - C ( c) By using `N = N_0((1)/(2))^(t//T)` , where `N = (1 - (7)/(8)) N_0 = (1)/(8) N_0` So `(1)/(8) N_0 = N_0((1)/(2))^(t//T) rArr ((1)/(2))^3 = ((1)/(2))^(t//5)` `rArr t = 15 days`. |
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| 243. |
A sample contains `16 gm` of radioactive material, the half-life of which is two days. After `32` days, the amount of radioactive material left in the sample isA. Less than `1 mg`B. `(1)/(4) gm`C. `(1)/(2) gm`D. 1 gm |
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Answer» Correct Answer - A (a) Remaining amount =`16 xx ((1)/(2))^(32//2) = 16 xx ((1)/(2))^16 = ((1)/(2))^12 lt 1 mg`. |
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| 244. |
The minimum frequency of a `gamma`-ray that causes a deutron to disintegrate into a poton and a neutron is `(m_(d)=2.0141 am u, m_p=1.0078 am u, m_n=1.0087 am u.)`.A. `2.7 xx 10^(20)Hz`B. `5.4 xx10^(20)Hz`C. `10.8 xx10^(20)Hz`D. `21.6xx10^(20)Hz` |
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Answer» Correct Answer - b Total mass of the products `=2.0165 a.m.u.`, which is greater than the mass of the deutron by `0.0024 a.m.u.` The extra mass must be provided by the energy of the photon so that minimum possible frequency must be given by `v=(0.0024 a.m.u. c^(2))/(h)" "(1 a.m.u. =1.66 xx 10^(-27) kg)` ` =5.4 xx 10^(20) Hz`. |
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| 245. |
The compound unstabel nucleus `._(92)^(236)U` often decays in accordance with the following reaction `._(92)^(236)U rarr ._(54)^(140)Xe +._(38)^(94)Sr ` + other particles During the reaction, the uranium nucleus 'fissions' (splits) into the two smaller nuceli have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons). Inside a nucleus, the nucleons (protonsa and neutrons)attract each other with a 'strong nuclear' force. All neutrons exert approxiamtely the same strong nuclear force on each other. This force holds the nuclear are very close together at intranuclear distances. A proton and a neutron are both shot at `100 m s^(-1)` toward a `._6^(12)C` nuleus. Which partilce, if either, is more likely to be absorebed by the nucleus?A. The protonB. The neutronC. Both particles are about equally likely to be absorbed,.D. Neither particle will be absorbed. |
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Answer» Correct Answer - b Once the neutron gets sufficiently close to the nucleus, the strong nuclear froces sucks it in. Same happens with the proton except it is electrostatically repelled by the six protons already inside the carbon nucleus. The repulsiohn prevents a `100 ms^(-1)` proton from getting close enough to the nucleus. Therefore, the answer is (b). |
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| 246. |
When an atom undergoes `beta^(bar)` decayA. a neutron changes into a protonB. a proton changes into a neutronC. a neutron changes into an antiprotonD. a proton changes into an antineutron |
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Answer» Correct Answer - b A nucleus contains protons and neutrons with no antiprotons and antineutrons. Hnece, answer can be either (b) or (d). Due to conservation of spin, the answer is (b) |
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| 247. |
A radioacitve nucleus is being produced at a constant rate `alpha` per second. Its decay constant is `lambda`. If `N_(0)` are the number of nuclei at time `t=0`, then maximum number of nuceli possible are .A. `alpha/lambda`B. `N_(0)+alpha/lambda`C. `N_(0)`D. `lambda/alpha+N_(0)` |
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Answer» Correct Answer - A |
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| 248. |
Of the following atoms `._(6)C^(14),._(7)N^(13),._(88)Ra^(236),._(7)N^(14),._(8)O^(16)` and `._(86)Rn^(232)` a pair of isobars is:A. `._(6)C^(11),._(7)N^(13)`B. `._(7)N^(13),._(7)N^(14)`C. `._(6)C^(14),._(7)N^(14)`D. `._(6)C^(14),._(8)O^(16)` |
| Answer» Correct Answer - C | |
| 249. |
Nuclei of radioactive element A are being produced at a constant rate. `alpha`. The element has a decay constant `lambda`. At time `t=0`, there are`N_(0)` nuclei of the element. (a) Calculate the number N of nuclei of A at time t. (b) IF `alpha =2 N_(0) lambda` , calculate the number of nuclei of A after one half-life time of A and also the limiting value of N at `t rarr oo`. |
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Answer» The rate of formation of radioactive nuclei is `alpha` . Rate of decay of radioactive nuclei is `lambda N`. Therefore, `(dN)/(dt) = alpha -lambda N` `rArr (dN)/(alpha -lambda N)=dt` Integrating, `(log e(alpha - lambda N_(0)))/(- lambda) =t +A` Where A is constant of integration. At `t =0,N =N_(0)` ` because (log_(e) (alpha lambda-lambdaN_(0)))/(-lambda)=A` (i) Therefore, Eq. (i) gives `(log_(e)(alpha-lambdaN))/(- lambda)=t+(log_(e)(alpha-lambdaN))/(-lamda)` `rArrlog_(e) ((alpha-lambdaN)/(alpha-lambdaN_(0)))=-lambdat` `rArr (alpha-lambdaN)/(alpha-lambdaN_(0))=e^(lambdat)` ltbtgt `rArr (alpha-lambdaN)/(alpha-lambdaN_(0))=e^(lambdat)` `rArr N=(alpha)/(lambda) (1 -e^(-lambda t)) + N_(0)e^(- lambda t)` (b) Given, `a = 2 N_(0) lambda` `t=T_(1)//2 =(0.693)/(lambda)` ` ` because N =1.5 N_(0)` When `t rarr infty`, Eq(ii) gives `N=(alpha)/(lambda).` |
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| 250. |
Consider the following nuclear decay: (initially `.^(236)U_(92)` is at rest) `._(92)^(236)rarr_(90)^(232)ThrarrX` Regarding this nuclear decay select the correct statement:A. The nucleus `X` may be at rest.B. The `._(90)^(232)Th` nucleus may be in excited state.C. The X may have kinetic energy but `._(90)^(232)Th` will be at restD. The `Q` valeus is `Delta mc^(2)` where `Delta mc^(2)` where `Delta m` is mass difference of `(._(92)^(236)U and . _(90)^(232))` and `c` is speed of light |
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Answer» Correct Answer - B (A) Since energy will be released `X` will not be at rest. (B) Generally daughter nucleus is in excited state. (C )If `X` has kinetic energy, `.^(232)Th` will also have kinetic energy to conserve the momentum (D) The `Q` value `Q=(m_(u)-m_(Th)-m_(x)).C^(2)` |
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