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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.
| 601. |
Consider the case of bombardment of `U^(235)` nucleus with a thermal neutron. The fission products are `Mo^(95)` & `La^(139)` and two neutrons. Calculate the energy released by one `U^(235)` nucleus. (Rest masses of the nuclides are `U^(235)=235.0439 u, ._(0)^(1)n=1.0087u, Mo^(95)=94.9058 u, La^(139).9061u)`. |
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Answer» Correct Answer - `[M_(U)+m_(n)-M_(La)-2m_(n)]931=207.9 MeV` `U^(235)+n rarrM_(0)^(95)+La^(139)+2n+Q` `Q=(m_(u)+m_(n)-m_(M_(0))-m_(La)-2m_(n)).c^(2)` `(235.0439+1.0087-94.9058-138.9061-2xx1.0087)xx931 MeV.` `207.9 MeV`. |
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| 602. |
For the `D-T` fusion reaction, find the rate at which deuterium & trithium are consumed to produce `1 MW`. The `Q`-value of `D-T` reactions is `17.6 MeV` & assume all the energy from the fusion rection is available. |
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Answer» Correct Answer - `(2)/(N_(A))xx(1)/(17.6e)xx10^(-3) Kg//s= 1.179xx10^(-9)kg//s` `(3)/(N_(A))xx(1)/(17.6e)xx10^(-3)kg//s = 1.769xx10^(-9) kg//s` Each deiterium nucleus prodeuces `17.6 MeV`. `1 kg` of deuterium `=(1xx10^(3))/(2)N_(A)` no. of deuterium `-=17.6xx(10^(3))/(2)N_(A) MeV` energy produced. `:.` To produce `1 MW`, amount of deuterium in `kg` required per second. `=(1xx10^(6))/(17.6xx(10^(3))/(2)xxN_(A)xxexx10^(6))kg//s=17.69xx10^(-9)kg//s` |
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| 603. |
Assuming that about `20 M eV` of energy is released per fusion reaction `._(1)H^(2)+._(1)H^(3)rarr._(0)n^(1)+._(2)He^(4)`, the mass of `._(1)H^(2)` consumed per day in a future fusion reactor of powder `1 MW` would be approximatelyA. `0.001g`B. `0.1g`C. `10.0g`D. `1000 g` |
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Answer» Correct Answer - b `P=10^(6)W` Time `= 1 day =24 xx 36 xx 10^(2)s` Energy produced, `U=Pt=10^(6) xx24 xx 36xx 10^(2)` `=24 xx 36 xx10^(8) J` Energy released per fusion reaction is `20 MeV =32 xx 10^(-13) J` Energy released per atom of `._(1)H^(2)` is `32 xx 10^(-13) J` Number of `._1H^2` atoms used is `(24 xx 36 xx10^(8))/(32 xx 10^(-12)=22 xx 10^(21))` Mass of `6 xx10^(23)` atom `=2g` Mass of `27 xx10^(21)` atoms` =(2)/(6) xx 10^(23) xx 27xx 10^(21) =0.1g.` |
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| 604. |
Assuming that about `20 M eV` of energy is released per fusion reaction `._(1)H^(2)+._(1)H^(3)rarr._(0)n^(1)+._(2)He^(4)`, the mass of `._(1)H^(2)` consumed per day in a future fusion reactor of powder `1 MW` would be approximatelyA. `0.1 gm`B. `0.01 gm`C. `1 gm`D. `10 gm` |
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Answer» Correct Answer - A no of moles of `._(1)H^(2)` consumed `=(1MWxx(24xx3600)sec//day)/((20 MeVxx6.023xx10^(23)))=0.05` `:. m=0.1 g` |
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| 605. |
Nuclei `X` decay into nuclei `Y` by emitting `alpha`-aprticles. Energies of `alpha`-particle are found to be only `1 MeV` & `1.4 MeV`. Disregarding the recoil of nuclei `Y`. The energy of `gamma` photon emitted will beA. `0.8 MeV`B. `1.4 MeV`C. `1 MeV`D. `0.4 MeV` |
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Answer» Correct Answer - D Energy of `gamma` photon `=` Difference in energies of `alpha` particles `=0.4 MeV` |
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| 606. |
Calculate the amount of energy released during the `alpha`-decay of `._(92)^(238)Urarr_(90)^(234)Th+._(2)^(4)He` Given: atomic mass of `._(92)^(238)U=238.05079 u`, atomic mass of `._(90)^(234)Th=234.04363 u`, atomic mass `._(2)^(4)He=4.00260u , 1u=931.5 MeV//c^(2)`. Is this decay spontaneous?Give reason. |
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Answer» The energy released in the `alpha`-decay is `Q= [m(._(92)^(238)U)-m(._(90)^(234)Th)-m(._(2)^(4)He)]c^(2)` `=[238.0579-234.04363-4.00260]xx931.5 MeV` `=0.00456xx931.5=4.25 MeV` As the `Q`-value is positive, the decay process is spontaneous. |
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| 607. |
`U^238` decays into `Th^234` by the emission of an a-particle. There follows a chain of further radioactive decays, either by `alpha`-decay or by `beta-`decay. Eventually a stable nuclide is reached and after that, no further radioactive decay is possible. which of the following stable nuclides is the end product of the `U^238` radioactive decay chain ?A. `Pb^206`B. `Pb^207`C. `Pb^208`D. `Pb^209` |
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Answer» Correct Answer - A (a) `(4n + 2)` series starts from `U^238` and its stable end product if `Pb^206`. |
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| 608. |
Stationery nucleus `.^(238)U` decays by a emission generaring a total kinetic energy T: `._(92)^(238) rarr ._(90)^(234)Th +._2^4 alpha` What is the kinetic energy of the `alpha`-particle?A. slightly less than `T`B. `T//2`C. slightly less than `T`D. slightly greater than `T` |
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Answer» Correct Answer - C ( c) Let then kinetic energy of the `alpha-`particle be `E_alpha` and that of the thorium Th be `E_(Th)` The ratio of kinetic energies is `(E_alpha)/(E_(th)) = ((1)/(2) m_alpha v_alpha^2)/((1)/(2) m_(th) v_(th)^2) = ((m_alpha)/(m_(th))) ((v_alpha)/(v_(th)))^2` ....(1) By conservation of momentum the momentum of `alpha-`particle and that of the recoiling thorium must be equal. Thus, `m_alpha v_alpha = m_(th) v_(th)` or `(v_(alpha))/(v_(th)) = (m_(th))/(_(alpha))` ...(2) Subst. (2) into (1), we have `(E_(alpha))/(E_(th)) = ((m_alpha)/(m_(th)))((m_(th))/(m_(alpha)))^2 = (m_(th))/(m_(alpha)) = (234)/(4) = 58.5` Thus, the kinetic energy is the `alpha-`particle expressed as the fraction of the total kinetic energy `T` is given by `E_alpha = (58.5)/(1 + 58.5) T = (58.5)/(59.5) T = 0.98 T` which is slightly less than `T`. |
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| 609. |
Stationery nucleus `.^(238)U` decays by a emission generaring a total kinetic energy T: `._(92)^(238) rarr ._(90)^(234)Th +._2^4 alpha` What is the kinetic energy of the `alpha`-particle?A. Slightly less than `T//2`B. `T//2`C. Slightly less than TD. Slightly greater than T |
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Answer» Correct Answer - c Let the kinetic energy of the `alpha`-particle be `E_(alpha)` and that of the thorium Th be `E_(th)`. The ratio of kinetic energies is `(E_(alpha))/(E_(th))=((1)/(2)m_(alpha)v_(alpha)^(2))/((1)/(2) m_(th)v_(th)^(2))=((m_(alpha))/(m_(th)))((v_(alpha))/(v_(th)))^(2)` By conservation of momentum, the momentum of `alpha`-particle and that of the recoiling thorium must be equal. Thus, `m_(alpha) v_(alpha)=m_(th) v_(th)` or `(v_(alpha))/(v_(th)) =(m_(th))/(m_(alpha))` Substitutuing Eq, (ii) in Eq. (i), we have `E_(alpha)/(E_(th)) =((m_(alpha))/(m_(th)))((m_(th))/(m_(alpha)))^(2) =(m_(th))/(m_(a))=(234)/(4)=58.5` Thus, the kinetic energy of the `alpha`-particle expressed as the fraction of the total kinetic energy `T` is given by `E_(alpha)=(58.5)/(1+58.5)T=(58.5)/(59.5)T=0.98T` Which is slightly less than `T`. |
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| 610. |
If `m,m_n` and `m_p` are masses of `._Z X^A` nucleus, neutron and proton respectively.A. `m lt (A - Z)m_n + Zm_p`B. `m = (A - Z)m_n + Zm_p`C. `m = (A - Z)m_p + Zm_n`D. `m gt (A - Z) m_n + Zm_p` |
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Answer» Correct Answer - A (a) The mass of nucleus formed is always less than the sum of the masses of the constituent protons and neutrons, i.e., `m lt (A - Z) m_n + zm_p`. |
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| 611. |
`M_n` and `M_p` represent mass of neutron and proton respectively. If an element having atomic mass `M` has `N-`neutron and `Z`-proton, then the correct relation will be :A. `Mlt[NM_(n)+ZM_(P)]`B. `Mgt[NM_(n)+ZM_(P)]`C. `M=[NM_(n)+ZM_(P)]`D. `M=N[M_(n)+M_(P)]` |
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Answer» Correct Answer - A |
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| 612. |
`M_p` denotes the mass of a proton and `M_n` that of a neutron. A given nucleus, of binding energy `B`, contains `Z` protons and `N` neutrons. The mass `M(N,Z)` of the nucleus is given by.A. `M(N,Z)=NM_(n)+ZM_(P)-Bc^(2)`B. `M(N,Z)=NM_(n)+ZM_(P)+Bc^(2)`C. `M(N,Z)=NM_(n)+ZM_(P)-B//c^(2)`D. `M(N,Z)=NM_(n)+ZM_(P)+B//c^(2)` |
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Answer» Correct Answer - C |
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| 613. |
If `m,m_n` and `m_p` are masses of `._Z X^A` nucleus, neutron and proton respectively.A. `M=(A-Z)m_(n)+Zm_(p)`B. `M=Zm_(n)+(A-Z)m_(p)`C. `M lt (A-Z) m_(n)+Zm_(p)`D. `M gt (A-Z) +Zm_(p)` |
| Answer» Correct Answer - A | |
| 614. |
`M_p` denotes the mass of a proton and `M_n` that of a neutron. A given nucleus, of binding energy `B`, contains `Z` protons and `N` neutrons. The mass `M(N,Z)` of the nucleus is given by.A. `m(N, Z)=Nm_(n)+ZM_(p)-Bec^(2)`B. `m(N, Z)=Nm_(n)+Zm_(p)+Bec^(2)`C. `m(N, Z)=Nm_(n)+Zm_(p)-BE//c^(2)`D. `m(N, Z)=Nm_(n)+Zm_(p)+BE//c^(2)` |
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Answer» Correct Answer - C |
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| 615. |
Which of the following is in the increasing order for penetrating power ?A. `alpha, beta, gamma`B. `beta, alpha, gamma`C. `gamma, alpha, beta`D. `gamma, beta, alpha` |
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Answer» Correct Answer - A (a) Penetration power of `gamma` is `100` times of `beta`, while that of `beta` is `100` times of `alpha`. |
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| 616. |
`M_n` and `M_p` represent mass of neutron and proton respectively. If an element having atomic mass `M` has `N-`neutron and `Z`-proton, then the correct relation will be :A. `M lt [NM_n + ZM_p]`B. `M gt [NM_n + ZM_p]`C. `M = [NM_n + ZM_p]`D. `M = N[M_n + M_p]` |
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Answer» Correct Answer - A (a) Given, Mass of neutron = `M_n` mass of proton `= M_p` , atomic mass of the element `= M` , number of neutrons in the element `= N` and number of protons in the element `= Z`. We know that the atomic mass `(M)` of any stable nucleus is always less than the sum of the masses of the constituent particles. Therefore, `M lt [NM_n + ZM_p]`. |
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| 617. |
A rock sample has radioactive isotope A that decays into a stable product B. half life of A is `10^(3)` year. It was found that the rock sample contained `n_(1)` moles of A and `(n_(1))/(4)` moles of b. Plot the variation of population of B with time. After how much time the rock will have equal population of A and B? [Given `2.^(0.68)=1.6`] |
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Answer» Correct Answer - 680y |
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| 618. |
A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life of one species is `tau` and that of the other is `5 tau`. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figure best represents the form of this plot? (a), (b), (c), (d) A. B. C. D. |
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Answer» Correct Answer - d `N_(1)=N_(0)e^(-t/tau)" "(i) and tau=(1)/(lambda)` `N_(2)=N_(0) e^(-lambda_(2) t) =N_(0) e^(- t/5tau)" "and 5 tau (1)/(lambda_(2))`. Adding (i)and (ii), we get `N=N_(1)+N_(2)=N_(0)(e^(-t//5tau) +e^(-t//5tau))` (a) is not correct option as there is a time `tau` for which `N` is constant, which means for time `tau` there is no process of radioactive atoms which is impossible as N will only decreases exponentially. Hence, the correct option is (d). |
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| 619. |
The activity of a radioactive sample is `1.6` curie, and its half-life is `2.5 days`. Its activity after `10 days` will beA. `0.8 curie`B. `0.4 curie`C. `0.1 curie`D. `0.16curie` |
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Answer» Correct Answer - c Since four half-lives have elapesed `A=(A_(0))/(2^(4)) =(A_(0))/(16)=(1.6)/(16) curie=0.1 curie`. |
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| 620. |
A radioactive sample consists of two isotopes one of then decays by `alpha-` emission with a half life of `tau_(1)=405s` and the other one decays by `beta` emission with half life of `tau_(1)`=1620s`. At time t=0, probabilities of getting `alpha` and `beta` particles from the sample are equal. (a) If a particle coming out of the sample is detected at t=1620s, what is the probability that it is `alpha` particle? (b) Find the time when total number of surviving nuclei is half the initial quantity. Given If `x^(4)+4x-2.5=0` then `x=0.59` `"log"_(10)2=0.301, "log"_(10)5.9=0.774` |
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Answer» Correct Answer - (a) `(1)/(9)` (b) 1215s |
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| 621. |
Radon `220` decays to Bismuth `212` by the following series of decay `{:(._(86)Rnrarr_(84),Po^(216)+_(2),He^(4),,T_(1//2)=55s,),(._(84)Po^(216)rarr_(82),Pb^(212)+_(2),He^(4),,T_(1//2)=0.016s,),(._(82)Pb^(212)rarr_(83),Bi^(212)+_(-1),e^(0),,T_(1//2)=10..6s,):}` If certain mass of randon is allowed to decay in certain container, after five minutes element with greatest and least mass will respectively beA. Randon, bismuthB. Polonium, leadC. Lead, bismuthD. Bismuth, lead |
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Answer» Correct Answer - C small `T_(1//2)` indicates that more such reactions take place. |
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| 622. |
Radon `(Ra)` decays into Polonium `(P_0)` by emitting an `alpha-`particle with half-life of `4` days. A sample contains `6.4 xx 10^10` atoms of `R_n`. After `12` days, the number of atoms of `R_n` left in the sample will beA. `3.2xx10^(10)`B. `0.53xx10^(10)`C. `2.1xx10^(10)`D. `0.8xx10^(10)` |
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Answer» Correct Answer - D |
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| 623. |
The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `A. 3.8 daysB. 16.5 daysC. 33 daysD. 76 days |
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Answer» Correct Answer - B (b) By formula `N = N_0 e^(-lamda t)` Given `(N)/(N_0) = (1)/(20)` and `lamda = (0.6931)/(3.8)` `rArr 20 = e^((0.6931)/(3.8))` Taking log of both sides or `log 20 = (0.6931 xx t)/(3.8) log_10 e` or `1.3010 = (0.6931 xx t xx 0.4343)/(3.8)` `rArr t = 16.5 days`. |
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| 624. |
Radon `(Ra)` decays into Polonium `(P_0)` by emitting an `alpha-`particle with half-life of `4` days. A sample contains `6.4 xx 10^10` atoms of `R_n`. After `12` days, the number of atoms of `R_n` left in the sample will beA. `3.2 xx 10^10`B. `0.53 xx 10^10`C. `2.1 xx 10^10`D. `0.8 xx 10^10` |
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Answer» Correct Answer - D (d) In the given case, `12` days = `3` half-lives Number of atoms left after `3` half-lives, =`6.4 xx 10^10 xx (1)/(2^3) = 0.8 xx 10^10`. |
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| 625. |
The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `A. `3.8 days`B. `16.5 days`C. `33 days`D. `76 days` |
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Answer» Correct Answer - b `T_(1//2)=3.8 day` `:. lambda =(0.693)/(t_(1//2))=(0.693)/(3.8)=0.182` If the initial number of atoms is a =`A_(0)`, then after time t the number of atoms is `a//20`=A. We have to find `t`. ` t=(2.303)/(lambda) log.(A_(0))/(A)=(2.303)/(0.182) log.(a)/(a//20)=(2.303)/(0.182) log.20` `=16.46 day`. |
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| 626. |
The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) ` |
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Answer» We know that `lambda =0.693//T_(1//2)` Here, `T_(1//2) =3.8 day` `:. Lambda =(0.693)/(3.8)=0.182 per day` If initially `(at t=0)` the number of atoms present be `N_(0)`, then the number of atoms N left after a time given by `N =N_(0) e^(-lambda t)` `(N)/(N_(0))= e^(-lambda t)` or `(1)/(20) =e^(- lambda t)` Taking log, `lambda t=log_(e) 20 =2.3026 log_(10) 20` `:. t=(2.3026 log_(10)20)/(lambda) = (2.3026 log_(10)20)/(0.182) =16.45 days`. |
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| 627. |
The half-life of radon is `3.8` days. Three forth of a radon sample decay in.A. `5.02` daysB. `15.2` daysC. `7.6` daysD. `11.4` days |
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Answer» Correct Answer - C ( c) Decayed fraction `= (3)/(4)`, so undecayed fraction `= (1)/(4)` Now `(N)/(N_0) = ((1)/(2))^n rArr (1)/(4) = ((1)/(2))^n rArr n = 2` `rArr t = n xx T_(1//2) = 2 xx 3.8 = 7.6 days`. |
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| 628. |
What is the wavelenth of the `0.186 MeV` gamma- ray photon emitted by radium `._(88)^(226) Ra`? |
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Answer» The photon energy is the difference between two nuclear energy levels. Equation `E_(i )- E_(f) =hf` gives the relation between the energy level separation `Delta E` and the frequency f of the photon as `Delta E =hf` . Since `f lambda =c` , the wavelength of the photon is `lambda -hc//Delta E`. First, we must convert the photon energy into joule: `Delta E =(0.186 xx 10^(6) eV)((1.60 xx 10^(-19))/1 eV)` ` = 2.98 xx 10^(-14) J` The wavelength of the photon is `lambda=(hc)/(Delta) E =((6.63 xx 10^(-34) J.s)(3.00 xx 10^(8) m s^(-1)))/(2.98 xx 10^(-14)J)` `=6.67+10^(-12)m`. |
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| 629. |
Uranium ores contain one radium `-226` atom for every `2.8 xx 106` uranium `-238` atoms. Calculate the half-life of `._92 U6238` given that the half-life of `._88 Ra^226` is `1600` years `(._88Ra^226` is a decay product of `._92 U^238`) :A. `1.75 xx 10^3 years`B. `1600 xx (238)/(92) years`C. `4.5 xx 10^9 years`D. `1600 years` |
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Answer» Correct Answer - C ( c) `N_1 lamda_1 = N_2 lamda_2` `T = (0.693)/(lamda)` Hence `2.8 xx 10^6 xx (0.693)/(T_1(U)) = 1 xx (0.693)/(T_2 (Ra))` `:. T_1 (U) = 1600 xx 2.8 xx 10^6` `=4.48 xx 10^9` years `~~ 4.5 xx 10^9` years. |
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| 630. |
The half-life of `Ra^226` is 1602 y.Calculate the activity of 0.1g of `RaCl_2` in which all the radium is in the form of `Ra^226`.Taken atomic weight of Ra to be the 226 g `mol^-1` and that of C1 to be `35.5 g mol^-1`. |
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Answer» Correct Answer - `[2.8xx10^(9) dps]` |
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| 631. |
The half-life of radium is `1620 years` and its atomic weight is `226`. The number of atoms that will decay from its `1 g` sample per second will be .A. `3.61xx10^(10)`B. `3.6xx10^(12)`C. `3.11xx10^(15)`D. `31.1xx10^(15)` |
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Answer» Correct Answer - A |
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| 632. |
The half-life of the radioactive nucleus `._(86)^(226)Ra` is `1.6 xx10^(3)yr`. If a sample contains`3.0 xx10^(16)` such nuclei, determine the activity at this time. |
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Answer» First, let us convert the half -life to seconds. `T_(1//2)=(1.6xx10^(3) years) (3.16xx16xx10^(7)s//years)` `=5.0 xx 10^(10) s` `because lambda=(0.693)/(T^(1//2))=(0.693)/(5.0xx10^(10)s)=1.4 xx10^(-11)s^(-1)` we case calculate the activity of the sample at `t=0` using `R_(0) =lambda N_(0)` , where `R_(0)` is the decay rate at` t=0` and `N_(0)` is the number of radioactive nuclei present at `t=0`. `R_(0) = lambda N_(0)=(1.4 xx 10^(-11) s^(-1))(3.0 xx 10^(16))` `=4.1 xx10^(5) decays s^(-1)` Beacuse `Ci=3.7 xx10^(10) decays s^(-1)` , the activity or decay rate at `t=0` is `R_(0) =11.1 mu Ci` . |
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| 633. |
If a nucleus such as `.^(226)Ra` that is initially at rest undergoes `alpha`- decay , then which of the following statemnets is true?A. The alpha particles has more kinetic than the daughter nuclues.B. The alpha particle has less kinetic energy than the daughter nucleus.C. The alpha particle and daughter nucleus both have same kinetic energyD. We cannot say anything about kinetic energy of alpha particle and daughter nucleus. |
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Answer» Correct Answer - a As the alpha particle decays, the daughter nulcues racoils. In such a process,the momentum conservation holds good. So, `P_(alpha) =P_(D) =P` `K_(alpha) =(P^(2))/(2 M_(alpha))` and `K_(D) =(p^(2))/(2 M_(D))` As `M_(D) gt M_(alpha)`, so `K_(alpha) gt K_(D)`. |
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| 634. |
Plutonium decays with a half-life of `24000` years. If the plutonium is stored for `72000` years, then the fraction of plutonium that remains is. |
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Answer» `T_(1//2)=24,000 "years"` Duration of time `(t)-= 72,000 "years"` Number of half lifes `(n)=(t)/(T_(1//2))=(72000)/(24000)=3` `:. 1g overset(1)rarr(1)/(2)g overset(2)rarr(1)/(4)goverset(3)rarr(1)/(g)g` `:.` Fraction of plutonium remains `=(1)/(8) g` |
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| 635. |
An alpha particle of energy `5 MeV` is scattered through `180^(@)` by a found uramiam nucleus . The distance of closest approach is of the order ofA. `1 A`B. `10^(-10)cm`C. `10^(-12)`cmD. `10^(-15)` cm |
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Answer» Correct Answer - C |
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| 636. |
In the Rutherford experiment, `alpha`-particles are scattered from a nucleus as shown. Out of the four paths, which path is not possible? A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - C |
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| 637. |
When the particle and its antiparticle unite, the result isA. a heavier particleB. two or more smaller particlesC. photonsD. partly matter and partly photons. |
| Answer» Correct Answer - C | |
| 638. |
The age of the wood if only `1//16` part of original `C^(14)` is present in its piece is (in years) ( `T` of `C^(14)` is `5,580` years)A. `5580`B. `11,160`C. `22320`D. `16740` |
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Answer» Correct Answer - C `n=4` half lifes |
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| 639. |
The `C^(14)` to `C^(12)` ratio in a certain piece of wood is `25%` of that in atmosphere. The half life period of `C^(14)` is `5,580` years. The age of wood piece is ( in years)A. `5,580`B. `2790`C. `1395`D. `11,60` |
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Answer» Correct Answer - D `N=N_(0)e^(-lambda t)` |
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| 640. |
The ratio of the amounts of energy released as a result of the fusion of `1 kg` hydrogen `(E_(1))` and fission of `1 kg` of `._(92)U^(236)(E_(2))` will beA. 1.28B. 3.28C. 5.28D. 7.28 |
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Answer» Correct Answer - D Energy released for nucleon in fission `=728` (energy released per nucleon in fission) `E_(1)=7.28 E_(2)` |
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| 641. |
A piece of wood is found to have the `(C^(14))/(C^(12))` ratio to be `0.5` times of that in a living plant The number of years back the plant died will be ( `T` of `C^(14)=5,580` years)A. `2,790` yearsB. `5,580` yearsC. `11,60` yearsD. `27,900` years |
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Answer» Correct Answer - B `t=1` half lifes |
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| 642. |
In the disintegration series `._(92)^(238)U overset rarr (alpha) X overset rarr (beta^(-))._Z^(A)Y` the values of `Z` and `A`, respectively, will beA. 92236B. 88230C. 90234D. 91234 |
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Answer» Correct Answer - D (d) `._92U^238 rarr ._2He^4 + ._90 X^234 rarr ._(-1)e^0 + ._91 U^234` Hence, `A = 234, Z = 91`. |
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| 643. |
In the disintegration series `._(92)^(238)U overset rarr (alpha) X overset rarr (beta^(-))._Z^(A)Y` the values of `Z` and `A`, respectively, will beA. `92326`B. `88230`C. `90234`D. `91234` |
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Answer» Correct Answer - d `alpha`-decay decreases mass number by `4` and reduces charge number by `2`. `beta`-deacy keeps mass number unchanged and increases charege by `1`. clearly, option (d) is the right choice . |
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| 644. |
Atomic mass number of an element is `232` and its atomic number is `90`. The end product of this radiaoctive element is an isotope of lead (atomic mass `208` and atomic number `82`.) The number of `alpha`-and `beta` -particles emitted are.A. `alpha = 3, beta = 3`B. `alpha = 3, beta = 4`C. `alpha = 6, beta = 0`D. `alpha = 4, beta = 6`. |
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Answer» Correct Answer - B (b) Number of `alpha-`particles emitted `= (232 - 208)/(4) = 6` Decrease in charge number due to `alpha-`emission `= 12` But decrease in charge number `=(90 - 82)` i.e., `8` Clearly, `4 beta-`particles are emitted. |
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| 645. |
Atomic mass number of an element is `232` and its atomic number is `90`. The end product of this radiaoctive element is an isotope of lead (atomic mass `208` and atomic number `82`.) The number of `alpha`-and `beta` -particles emitted are.A. `6,3`B. `6,4`C. `5,5`D. `4,6` |
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Answer» Correct Answer - b Decreases in mass number `=232 -208 =24` Number of `alpha`-particles emitted `=(24)/(4=6)` Due to emission of 6 particles, decreases in charge number is `12`. But actual decrease in number is `8`. Clearly, 4 `beta`- particles are emitted. |
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| 646. |
Atomic mass number of an element is `232` and its atomic number is `90`. The end product of this radiaoctive element is an isotope of lead (atomic mass `208` and atomic number `82`.) The number of `alpha`-and `beta` -particles emitted are.A. `alpha=3,beta=3`B. `alpha=6,beta=4`C. `alpha=6,beta=0`D. `alpha=4,beta=6` |
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Answer» Correct Answer - B |
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| 647. |
Half lives of two isotopes X and Y of a material are known to be `2xx10^(9)` years and `4xx10^(9)` years respectively if a planet was formed with equal number of these isotopes, then the current age of planet, given that currently the material has `20%` of X and `80%` of Y by number, will be:A. `2xx10^(9)` yearsB. `4xx10^(9)` yearsC. `6xx10^(9)` yearsD. `8xx10^(9)` years |
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Answer» Correct Answer - D |
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| 648. |
A good moderator shouldA. be a gasB. have appetite for neutronsC. be lighter in mass numberD. heavier in mass number |
| Answer» Correct Answer - C | |
| 649. |
Who designed the atomic reactor ?A. WilsonB. FermiC. RutherfordD. Teller |
| Answer» Correct Answer - B | |
| 650. |
In nuclear reactions, we have the conservation ofA. mass number and energyB. mass number and charge numberC. charge number and massD. mass number, charge number and energy |
| Answer» Correct Answer - D | |