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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.
| 301. |
A radioactive element has half-life period `800 yr`. After `6400 yr`, what amount will remain?A. `1/2`B. `1/16`C. `1/8`D. `1/256` |
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Answer» Correct Answer - D |
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| 302. |
The nucleus `._48^115 Cd` after two successive `beta^-` decays will give.A. `._46^115 Pa`B. `._49^114 In`C. `._50^113 Sn`D. `._50^115 Sn` |
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Answer» Correct Answer - D (d) `._48 Cd^115 overset (2(._-1beta^o))rarr ._50 Sn^115`. |
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| 303. |
The nucleus `._48^115 Cd` after two successive `beta^-` decays will give.A. `._(46)Pa^(115)`B. `._(49) ln^(114)`C. `._(50)Sn^(113)`D. `._(50)Sn^(115)` |
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Answer» Correct Answer - D |
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| 304. |
What is the radius of iodine atom (at no. `53`, mass number `126`)?A. `2.5xx10^(-11) m`B. `2.5 xx10^(-9) m`C. `7xx10^(-9) m`D. `7xx10^(-6) m` |
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Answer» Correct Answer - A |
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| 305. |
What is the radius of iodine atom (at no. `53`, mass number `126`)?A. `2.5xx10^(-11)m`B. `2.5xx10^(-9)m`C. `7xx10^(-9)m`D. `7xx10^(-6)m` |
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Answer» Correct Answer - A |
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| 306. |
Radius of `._2^4 He` nucleus is `3` Fermi. The radius of `._82^206 Pb` nucleus will be.A. 5 FermiB. 6 FermiC. 11.16 FermiD. 8 Fermi |
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Answer» Correct Answer - C |
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| 307. |
In the option given below , let `E` denote the rest mass energy of a nucleas and `n` a neutron .The correct option isA. `E(._(92)^(236)U) gt E(._(53)^(137)I) + E(._(39)^(97)Y) + 2E(n)`B. `E(._(92)^(236)U) lt E(._(53)^(137)I) + E(._(39)^(97)Y) + 2E(n)`C. `E(._(92)^(236)U) lt E(._(56)^(140)Ba) + E(._(36)^(94)Kr) + 2E(n)`D. `E(._(92)^(236)U) = E(._(56)^(140)Ba) + E(._(36)^(94)Kr) + 2E(n)` |
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Answer» Correct Answer - a Iodine and Yttrium are medium sized nuclei and therefore have more binding energy per nucleon as compared to uranium which has a big nuclei and less BE/nucelon. In other words, Iodine and Y ttrium are more stable and therefore possess less energy and less rest mass. Also, when uranium nuclei explodes, it will convert inti I and Y nuclei having kinetic energies. |
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| 308. |
A stable nuclei C is formed from two radioactive nuclei A and B with decay constant of `lambda_1` and `lambda_2` respectively. Initially, the number of nuclei of A is `N_0` and that of B is zero. Nuclei B are produced at a constant rate of P. Find the number of the nuclei of C after time t. |
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Answer» Correct Answer - `N_(0)(1-e^(-lambda_(1)t))+P(t+(e^(-lambda_(2)t-1))/(lambda))` `(dN_(B))/(dt) =P-lambda_2 N_B` `int_(0)^(N_2) (dN_(B))/(P)-lambda_(2) (N_(B))/(P)=int_(0)^(t)dt` `ln((P-lambda_(2) (N_(B)))/(P))-lambda_2t` `N_B=P(1 -e^(-lambda_(2)t)` The number of nuclei of A after time `t` is `N_(A)=N_(0) e^(-lambda t)`. Thus, `(dN_(C))/(dt) =lambda_(1)N_(A) +lambda_(2)N_(B)` `=+lambda_(1)N_(0) e^(-lambda t)+P(1-e^(-lmabda_(2) t)` `N_(C)+N_(0)(1-e^(- lmabda_1 t)) + P(t+(e^(-lambda_2 t-1))/(lambda_2))` |
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| 309. |
An excited `.^(69)Zoverset(*)n` Nucleus at rest decays to ground state `.^(69)Zn` by way of emitting a gamma photon. The energy equivalent of difference in mass of `Zoverset(*)n` and Zn is 0.44 MeV. Calculate the kinetic energy of recol of Zn nucleus. Rest mass of `.^(69)Zn` is `M_(0)=68.927u`. |
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Answer» Correct Answer - 1.5eV |
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| 310. |
An alpha active element has molar mass M and has a half life of `tau`. A large plate of thickness `alpha` is made from the radioactive element. Density of the plate is `rho`. Calculate the number of alpha particles emitted in unit time from unit surface area of the plate. Avogadro’s number is `N_(A)`. |
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Answer» Correct Answer - `(ln^(2)2)/(2)(alpharhoN_(A))/(tauM)` |
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| 311. |
`M_(x) and M_(y)` denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q - value for a `beta-` decay is `Q_(1)` and that for a `beta^(+)` decay is `Q_(2)`. If `m_(e)` denotes the mass of an electrons, then which of the following statements is correct?A. `Q_(1)(M_9x)-M_(y))` and `Q_(2)=[M_(x)-M_(y)-2m_(e )]c^(2)`B. `Q_(1)(M_9x)-M_(y))` and `Q_(2)=(M_(x)-M_(y))c^(2)`C. `Q_(1)=(M_(x)-M_(y)-2m_(e ))c^(2)` and `Q_(2)=(M_(x)-M_(y)+2c_(e ))c^(2)`D. `Q_(1)=(M_(x)-M_(y)+2m_(e ))c^(2)` and `Q_(2)=(M_(x)-M_(y)+2m_(e ))c^(2)` |
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Answer» Correct Answer - A `beta` decay, `_(z)X^(A)rarr_(z-1)A^(Y)+_(-1)e^(0)+bar(v)` `Q_(1)=[m_("nucleus")(._(z)X^(A))-m_("neucleaus")(._(z+1)Y^(A))-m_(e )]c^(2)` `:. M_("nucleus") = m_("atom") - z_(me)` `:. Q_(1) = [(m_(x) - z_(me)) - (m_(y) - (z+1) me)]c^(2)` `=(M_(x)-M_(y))C^(2)` Similarly in `beta^(+)` decay. `Q_(2)(M_(x)-M_(y)-2me)c^(2)` |
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| 312. |
If `10%` of a radioactive material decays in `5` days, then the amount of original material left after `20` days is approximately.A. 0.6B. 0.65C. 0.7D. 0.75 |
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Answer» Correct Answer - B (b) Disintegrating percentage `= 10` Surviving percentage `= 100 - 10 = 90` Surviving fraction `=f` `f = (90)/(100) = 0.9` `t_f = 5 days t = 20 days` `n_f = (t)/(t_f) = (20)/(5)= 4` `(N)/(N_0) = (f)^(n_f) = (0.9)^4` ` :. ((N)/(N_0)) xx 100 = 0.6560 + 100` `=65.6 %`. |
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| 313. |
`10 gm` of radioactive material of half-life `15` year is kept in store for `20` years, The disintegrated material is.A. `12.5 g`B. `10.5 g`C. `6.03 g`D. `4.03 g` |
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Answer» Correct Answer - C ( c) Remaining material `N = (N)/(2^(t//T))` `rArr N = (10)/((2)^(20//15)) = (10)/(2.15) = 3.96 gm` So decayed material `= 10 - 3.9 = 9.04 gm`. |
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| 314. |
Half-life of a radioactive substance is `20` minutes. Difference between points of time when it is `33 %` disintegrated and `67 %` disintegrated is approximate.A. 10 minB. 20 minC. 30 minD. 40 min |
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Answer» Correct Answer - B (b) `lamda = (0.693)/(T_(1//2)) = (0.693)/(20) = 0.03465` Now time od decay `t = (2.303)/(lamda) log ((N_0)/(N))` `rArr t_1 = (2.303)/(0.03465) log((100)/(67)) = 11.6 min` and `t_2 = (2.303)/(0.03465) log ((100)/(33)) = 32 min` Thus time difference between points of time =`t_1 - t_2 = 32 - 11.6 = 20.4 min ~~ 20 min`. |
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| 315. |
A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its `1 cm^2`window. If the source contains `6.0xx10^16`active nuclei and the counter records a rate of 50000 counts//second, find the decay constant. Assume that the source emits alpha particles fall nearly normally on the window. |
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Answer» Correct Answer - `[1.05 xx 10^(-7) secc^(-1)]` |
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| 316. |
`A` and `B` are two radioactive substances whose half lives are `1` and `2` years respectively. Initially `10 gm` of `A` and `1 gm` of `B` is taken. The time (approximate) after which they will have same quantity remaining is.A. 6.62 yearsB. 5 yearsC. 3.2 yearsD. 7 years |
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Answer» Correct Answer - A |
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| 317. |
A nuclear transformation is denoted by `X(n,alpha) rarr. _(3)^(7)Li`. Which of the following is the nucleus of element `X`A. `._(6)^(12)C`B. `._(5)^(10)B`C. `._(5)^(9)B`D. `._(4)^(11)Be` |
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Answer» Correct Answer - B `._(Z)^(A)X+._(0)^(1)nrarr._(3)^(7)Li+._(2)^(4)He` It implies that `A+1=7+4` `rArrA=10` and `Z+0=3+2` `rArr Z=5` Thus, it is Boron `._(5)^(10)B` |
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| 318. |
Which of the following statement (s) is (are) correct ?A. The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons.B. The rest mass of a stable nucleus is greater than the sum of the rest masses of its seprated nucleons.C. In nuclear fission, energy is released by fusing two nuclei of medium mass (approximately 100 u)D. In nuclear fission, energy is released by fragmentation of very heavy nucleus. |
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Answer» Correct Answer - A, D |
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| 319. |
The decay constant of a radioactive substance is 0.173 `("years")^(-1)`. Therefore:A. Nearly `63%` of the radioactive substance will decay in `(1//0.173)` yearB. Half life of the radio active substance is `(1//0.173)` years.C. One-fourth of the radioactive substance will be left after nearly 8 yearsD. All the above statement are true |
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Answer» Correct Answer - A, C |
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| 320. |
The decay constant of a radioactive substance is` 0.173 year^(-1)`. Therefore,A. Nearly `63%` of the radioactive substance will decay in `(1//0.173)` year.B. half-life of the radio active subtance is `(1//0.173)` year.C. one-forth of the radioactive substance will be left after nerarly `8` years.D. half of the substance will decay in one average life time. |
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Answer» Correct Answer - A::C Given , `lambda=0.173` `T_(1//2)=(ln2)/(lambda)=(0.693)/(0.173)~=4` Also `N_(0)-N=N_(0)e^(-lambdat)` For `t=(1)/(0.173)"year"` `N_(0)-N=(N_(0))/(e )=0.37N_(0)` |
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| 321. |
The `beta`-decay process, discovered around `1900`, is basically the decay of a neutron `(n)`, In the laboratory, a proton `(p)` and an electron `(e^(-))` are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a tro-body dcay process, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., `n rarr p + e^(-)+overset(-)v_(e )`, around `1930`, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(overset(-)V_(e ))` to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is `0.8xx10^(6)eV`. The kinetic energy carried by the proton is only the recoil energy. If the anti-neutrino has a mass of `3eV//c^(2)` (where `c` is the speed of light) instead of zero mass, what should be the range of the kinetic energy, `K` of the electron?A. `0 leK le0.8xx10^(6)eV`B. `3.0 eV le K le 0.8xx10^(6) eV`C. `3.0 eV le K lt 0.8xx10^(6)eV`D. `0 le K lt 0.8xx10^(6) eV` |
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Answer» Correct Answer - D `0 le KE_(beta^(-))leQ-KE_(P)-KE_(overset(-)v)` `0 le KE_(beta^(-)) ltQ` |
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| 322. |
The `beta - decay` process , discovered around `1900` , is basically the decay of a neutron `n`. In the laboratory , a proton `p` and an electron `e^(bar)` are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e. ` n rarr p + e^(bar) + bar nu _(e) , ` around `1930` , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(bar nu_(e))` to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is `0.8 xx 10^(6) eV` The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino ?A. The nucleus `._3^6Li` can emit an alpha particleB. The nucleus `._(84)^(120) P0` cam emit a proton.C. Deuteron and alpha particle can undergo complete fusion.D. The nuclei `._(30)^(70) Zn` and `._(34)^(82) Se` can undergo complete fusion. |
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Answer» Correct Answer - C `KE_max of beta^(bar)` `Q=0.8 xx 10^(6) eV` `KE_p +KE_(beta^(bar))+KE_(bar(V))=Q` `KE_p` is almost zero When `KE_(beta(bar)) =0` Then `KE_(bat(barv))=Q-KE~=Q` |
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| 323. |
The `beta - decay` process , discovered around `1900` , is basically the decay of a neutron `n`. In the laboratory , a proton `p` and an electron `e^(bar)` are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e. ` n rarr p + e^(bar) + bar nu _(e) , ` around `1930` , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(bar nu_(e))` to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is `0.8 xx 10^(6) eV` The kinetic energy carried by the proton is only the recoil energy. If the - neutrono had a mass of `3 eV// c^(2)` (where c is the speed of light ) insend of zero mass , what should be the range of the kinectic energy `K.` of the electron ? |
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Answer» Correct Answer - D `0 le KE_(beta^(bar)) le Q` `0 le KE_(beta^(bar)) lt Q` |
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| 324. |
To determine the half life of a radioactive element , a student plote a graph of in `|(dN(t))/(dt)| versus t , Here |(dN(t))/(dt)|` is the rate of radioatuion decay at time t , if the number of radoactive nuclei of this element decreases by a factor of p after `4.16 ` year the value of p is A. `5319`B. `5422`C. `5707`D. `5818` |
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Answer» Correct Answer - 8 `N =N_0 e^(-lambda t)` `ln|dN|dt|=1n(N_0 lambda)-lambda t` From graph `lambda =(1)/(2)` per year `(t_(1))/(2)=(0.693)/(1//2)=1.386` year `4.16` years `=3t_(1//2)` `p=8`. |
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| 325. |
The activity of a freshly prepared radioactive sample is `10^(10)` disintegrations per second , whose mean life is `10^(9)s` The mass of an atom of this radioisotope is `10^(-25) kg ` The mass (in mg) of the radioactive sample is |
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Answer» Correct Answer - 1 `N=N_0 e^(-lambda t)` `(dN)/(dt)=10^(10)=N_0(lambda)e^(-10^(-9)t)` At `(t=0)` `10^(10)=N_(0) 10^(-9)` Mass of sample `=N_0 =10^(-25)` =`N_0` (mass of the atom) `=10^(-6) kg` `=10^(-6) xx 10^(3) g` `=10^(-3) g` `=1 mg` . |
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| 326. |
The activity of a freshly prepared radioactive sample is `10^(10)` disintergrations per second, whose mean life is `10^(9)s`. The mass of an atom of this radioisotope is `10^(-25) kg`. The mass ( in `mg`) of the radioactive samples is |
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Answer» Correct Answer - `1` `N=N_(0)e^(-lambdat)` `(dN)/(dt)=10^(10)=N_(0)(lambda)e^(-10^(-9)t)` at `(t=0)` `10^(10)=N_(0)10^(-9)` `N_(0)=10^(19)` mass of sample `=N_(0)10^(-25)` `=10^(-6)kgm` `=10^(-6)xx10^(3)gm` `10^(-3)gm` `=1 mg` |
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| 327. |
Which of the following process represents a `gamma- decay`?A. `.^A X_Z + gamma rarr .^A X_(Z- 1) + a + b`B. `.^A X_Z + .^1 n_0 rarr .^(A - 3) X_(Z -2) + c`C. `.^A X_Z rarr .^A X_Z + f`D. `.^A X_Z + e._-1 rarr .^A X_(Z -1) + g` |
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Answer» Correct Answer - C ( c) In a gamma decay process. There is no change on either `A` to `Z`. |
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| 328. |
Which of the following process represents a `gamma- decay`?A. `^(A)X_Z +gamma rarr ^(A)X_(Z-1) +a+b`B. `^(A)X_Z +^(1)n_(0) rarr ^(A-3)X_(Z-2) +ac`C. `^(A)X_(Z) rarr ^(A)X_(Z) +f`D. `^(A)X_(Z) +e_(-1) rarr ^(A)X_(Z-1) +g` |
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Answer» Correct Answer - c In `gamma`-decay, the atomic number and mass number do not change. |
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| 329. |
For uranium nucleus how does its mass vary with volume?A. ` m prop V`B. ` m prop 1//V`C. ` m prop (sqrtV)`D. ` m prop V^(2)` |
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Answer» Correct Answer - a We know that radius of the nucleus, `R=R_(0) A^(1//3)` Where `A` is the mass number. `:. R^(3)=R_(0)^(3)A rArr (4)/(3) pi R^(3)=(4)/(3)piR_(0)^(3)A` `rArr` volume prop mass` |
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| 330. |
A nucleus with mass number `220` initially at rest emits an `alpha`-particle. If the Q-value of the reaction is `5.5 MeV`, calculate the kinetic energy of the `alpha`-particle. (a) 4.4 MeV (b) 5.4 MeV (c) 5.6 MeV (d) 6.5 MeVA. `4.4`B. `5.4`C. `5.6`D. `6.5` |
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Answer» Correct Answer - B `KE=[(A-4)/(A)]Q` |
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| 331. |
Free `^238 U` nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measues that the separation betwee the alpha particle and the recoiling nucleus becomes `x` in time `t` after the decay. If a decay takes place when the train is moving at a uniform speed `v`, the distance between the alpha particle and the recoiling nucleus at a time `t` after the decay, as measured by the passenger will beA. `x+upsilont`B. `x-upsilont`C. `x`D. depends on the direction of the train |
| Answer» Correct Answer - C | |
| 332. |
The amount of heat generated by `1.0 mg` of a `Po^(210)` preparation during the one mean lifetime period of there nuclei, if the emitted a-particles are known to possess the kinetic energy `5.3 MeV` is `n(0.2 MJ)`. Assume all daughter nuclei are partically formed in ground state. What is the value of `n` ? |
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Answer» Number of nuclei initially present `=(10^(-3))/(210)xx6.023xx10^(23)=2.88xx10^(18)` The no. of decay in one mean lifetime `=(1-(1)/(e))xx2.88xx10^(18)` Energy released `= 2.88xx10^(18)xx(1-(1)/(e))xx5.3xx1.6xx10^(-13)J` |
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| 333. |
`A` and `B` are isotopes. `B` and `C` are isobars. All three are radioactive. Which one of the following is true.A. `A,B` and `C` must belong to the same elementB. `A,B` and `C` may belong to the same elementC. It is possible that `A` will change to `B` through a radioactive decay processD. It is possible that `B` will change to `C` through a radioactive decay process |
| Answer» Correct Answer - D | |
| 334. |
The explosion of the atomic bomb takes place due toA. Nuclear fissionB. Nuclear fusionC. ScatteringD. Thermionic emission |
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Answer» Correct Answer - A (a) In atom bomb nuclear fission takes place with huge temperature. |
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| 335. |
A sample of radioactive material is used to provide desired doses of radiation for medical purposes. The total time for which the sample can be used will dependA. only on the number of times radiation is drawn form itB. only on the intensity of doses drawn from itC. on both (a) and (b)D. neither on (a) and (b) |
| Answer» Correct Answer - D | |
| 336. |
In the chemical analysis of a rock the mass ratio of two radioactive isotopes is found to be `100:1`. The mean lives of the two isotopes are `4xx10^9` years and `2xx10^9` years, respectively. If it is assumed that at the time of formation the atoms of both the isotopes were in equal propotional, calculate the age of the rock. Ratio of the atomic weights of the two isotopes is `1.02:1`. |
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Answer» `(m_(A))/(m_(B)) = (100)/(1),(M_(A))/(M_(B)) = (1.02)/(1)` `bar(T_(A))=4 xx10^(9) year, bar(T_(B)) = 2xx10^(9)year` At `t = 0, ((N_A)_(0))/((N_B)_(0)) =(1)/(1)` `t = t, (N_(A))/(N_(B)) = (m_(A)//M_(A))/(m_(B)//M_(B)) = (m_(A))/(m_(B)).(M_(B))/(M_(A))` `= (100)/(1)xx(1)/(1.02) = (100)/(1.02)` `lambda_(A) = (1)/(bar(T_(A))) = (1)/(4xx10^(9)) year^(-1), lambda_(B) = (1)/(2xx10^(9))year^(-1)` `N_(A) = (N_A)_(0)e^(-lambda_(A)t)` `N_(B) = (N_B)_(0)e^(-lambda_(B)t)` `(N_(A))/(N_(B)) = e(lambda_(B)-lambda_(A))t` `(100)/(1.02) = e^(((1)/(2xx100^(9))-(1)/(4xx10^(9)))t) =e^((t)/(4xx10^(9))` `ln((100)/(1.02)) = (t)/(4xx10^(9))` `t= 4 xx 10^(9) xx 4.58` `= 1.83 xx 10^(10)year` |
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| 337. |
A pair of isotopes isA. `._(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 - B | |
| 338. |
Atoms having the same …… but different ….. are called isotopes . |
| Answer» Atomic number ,mass number | |
| 339. |
When `_(3)Li^(7)` nuclei are bombarded by protons , and the resultant nuclei are `_(4)Be^(8)` , the emitted particle will beA. neutronsB. alpha particlesC. beta particlesD. gamma photons |
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Answer» Correct Answer - D Gamma-photon. |
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| 340. |
When `_(3)Li^(7)` nuclei are bombarded by protons , and the resultant nuclei are `_(4)Be^(8)` , the emitted particle will beA. alpha particlesB. beta particlesC. gamma photonsD. neutrons |
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Answer» Correct Answer - C `._(3)Li^(7)+_(1)H^(1)rarr._(4)Be^(8)+gamma` |
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| 341. |
There are two radioactive substance `A` and `B`. Decay consant of `B` is two times that of `A`. Initially, both have equal number of nuceli. After n half-lives of `A`,rates of disintegaration of both are equal. The value of `n` is .A. 1B. 2C. 4D. All of these |
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Answer» Correct Answer - A |
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| 342. |
Find the half-life of uranium, given that `3.32 xx 10^(7) g` radium is found per gram of uranium in old minerals. The atomic weights of uranium and radium are `238` and `226` and half-life of radium is `1600` years (Avogadro number is `6.023 xx 10^(23)//g -"atom"`). |
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Answer» In very old minerals, the amount of an elemant is constant. This implies that the element exists in radioactive equilibrium. Thus, here we can use `lambda_(U) N_(U) =lambda_(R)N_(R)` or `(N_(U))/(T_(U))=(N_(R))/(T_(R))` or `T_(u)=(N_(u))/(N_(R)) xxT_(R)` or `T_(U) =(m_(U) A_(R))/(m_(R)A_(U)) xx T_(R)` or `T_(U) = (1 xx 226)/(3.32 xx 10^(-7) xx 238) xx 1600 years` `=4.7 xx 10^(9) years` . |
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| 343. |
A certain substance decays to `1//32` of its initial activity in `25` days. Calculate its half-life.A. 1 dayB. 3 daysC. 5 daysD. 7 days |
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Answer» Correct Answer - C `(dN)/(dt)=lambda N=` avagadrono/226, `lambda=1//tau` |
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| 344. |
A certain substance decays to `1//32` of its initial activity in `25` days. Calculate its half-life. |
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Answer» `1goverset(1)rarr(1)/(2)goverset(2)rarr(1)/(4)goverset(3)rarr(1)/(8)g overset(4)rarr(1)/(16)g` `overset(5)rarr(1)/(32)g :. n=5` `(n) =(t)/(t_(1//2))rArrt_(1//2)=(t)/(n)=(25)/(5) , t_(1//2)=5 "days"` |
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| 345. |
(a) The rest mass energy of an electron is `m_(0)c^(2)=0.51MeV`. Can a photon of this energy create an electron in free space? (A) A single photon of energy `2m_(0)c^(2)` or greater has enough energy to form an electron `(e^(-))`. Positron `(e^(+))` pair. But prove that this process cannot occur in free space. (c) The pair production is possibel in presence of a third particle. Assume that a photon having frequency v collides with a nucleus of rest mass `M_(0)` (at rest) and creates an `e^(-),e^(+)` pair at rest. Calculate the mass `M_(0)` of the nucleus. Neglect relativistic effect in all questions. |
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Answer» Correct Answer - (a) No (b) `m_(0)=(h^(2)v^(2))/(2c^(2)[hv-2m_(0)c^(2)])` |
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| 346. |
Rate of evaporation (volume evaporated per unit time) of water kept in an earthen pot is proportional to the volume of water present in the pot. It is observed that it takes 48 hrs for 75% of the water kept in the pot to evaporate. The empty pot was placed below a tap from which water leaks in small drops each of volume `v_(0)`. The drops fall at a uniform rate of n drops per hour. Calculate the volume of water in the pot 24hrs after it was kept below the tap. |
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Answer» Correct Answer - `(12m_(0))/(ln2)` |
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| 347. |
A star initially has `10^(40) ` deuterons. It produces energy via the process `_(1)H^(2) + _(1)H^(2) + rarr _(1) H^(3) + p. `and `_(1)H^(2) + _(1)H^(3) + rarr _(2) He^(4) + n` .If the average power radiated by the state is `10^(16) W` , the deuteron supply of the star is exhausted in a time of the order of . The masses of the nuclei are as follows: `M(H^(2)) = 2.014 amu,` `M(p) = 1.007 amu, M(n) = 1.008 amu, M(He^(4)) = 4.001 amu`.A. `10^(6)s`B. `10^(8)s`C. `10^(12)s`D. `10^(16)s` |
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Answer» Correct Answer - C mass defect `._(Delta)m=0.026 "amu"` Energy released per reaction is `=0.026xx931.5 MeV=24.2 MeV` `=3.87xx10^(-12)J` Energy released by consuming `10^(40)` deuterons is `=3.87xx10^(-12)xx(10^(40))/(3)=12.9xx10^(28)J` Power=energy/time |
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| 348. |
A star initially has `10^(40) ` deuterons. It produces energy via the process `_(1)H^(2) + _(1)H^(2) + rarr _(1) H^(3) + p. `and `_(1)H^(2) + _(1)H^(3) + rarr _(2) He^(4) + n` .If the average power radiated by the state is `10^(16) W` , the deuteron supply of the star is exhausted in a time of the order of . The masses of the nuclei are as follows: `M(H^(2)) = 2.014 amu,` `M(p) = 1.007 amu, M(n) = 1.008 amu, M(He^(4)) = 4.001 amu`.A. `10^(6)` sB. `10^(8)` sC. `10^(12)` sD. `10^(16)` s |
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Answer» Correct Answer - C |
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| 349. |
A radioactive element `A` with a half-value period of `2` hours decays giving a stable element `Y`. After a time `t` the ratio of `X` and `Y` atoms is `1 : 7` then `t` is :A. 6 hoursB. 4 hoursC. between 4 and 5 hoursD. 14 hours |
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Answer» Correct Answer - A (a) After lapse of time `t`, let the number of atoms of element and `Y` element be respectivity `n_x` and `n_Y`. Then `(n_y)/(n_x) = 7` or `(n_y)/(n_x) + 1 = 7 +1` or `(n_y + n_y)/(n_x) = 8` or `(n_x)/(n_y +n_x) = (1)/(8)` Death of an atom of mother element means the birth of an atom of daughter element. `:. (n_x)/(n_y + n_x)(N)/(N_0) =(1)/(8) = ((1)/(2))^n` `:. n = 3 = (t)/(T)` But `T = 2 hours`, hence `t = 6 hours`. |
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| 350. |
`3.8` days is the half-life period of a sample. After how many days. The sample will become `1//8 th ` of the original substance ?A. `11.4`B. `3.8`C. 3D. None of these |
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Answer» Correct Answer - A (a) `(N)/(N_0) = ((1)/(2))^n rArr (1)/(8) = ((1)/(2))^n rArr n = 3` Now `t = n xx T_(1//2) = 3 xx 3.8 = 11.4 days`. |
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