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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.
| 251. |
Nuclei of a 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 . The number N of nuclei of A at time t isA. Number of nuclei of `A` at time `t` is `(1)/(lambda)[alpha-(alpha-lambdaN_(0))e^(-lambdat)]`B. Number of nuclei of `A` at time `t` is `(1)/(lambda)[(alpha-lambdaN_(0))e^(-lambdat)]`C. If `alpha=2N_(0)lambda`, then the limiting value of number of nuclei of `A (rarr prop)` will be `2N_(0)`D. If `alpha=2N_(0)lambda`, then the number of nuclei of `A` after one half-life of `A` will be `N_(0)//2` |
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Answer» Correct Answer - A::C Let at time `t`, number of radioactive nuclei are `N`. Net rate of formation of nuclei of `A` `(dN)/(dt)=alpha-lambdaN` or `(dN)/(alpha-lambdaN)=dt` `-(1)/(lambda)[l n(alpha-lambdaN)]_(No)^(N)=t` `l n(alpha-lambdaN)/(alpha-N_(0))= -lambdat` `N=(1)/(lambda)[alpha-(a-lambdaN_(0))e^(-lambdat)]` |
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| 252. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`). After a long time bombardment, number `.^(56)Mn` nuclei present in the target depends upon.A. All (a), (b) and (c ) are correctB. only (a) and (b) are correctC. only (b) and (c )are correctD. only (a) and (c ) are correct |
| Answer» Correct Answer - C | |
| 253. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`). At what constant rate P, `.^(56)Mn` nuclei are being produced in the cyclontron during the bombardment?A. `2xx10^(11) "nuclei"//s`B. `13.86xx10^(10) "nuclei"//s`C. `9.6xx10^(10) "nuclei"//s`D. `6.93xx10^(10) "nuclei"//s` |
| Answer» Correct Answer - B | |
| 254. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`). At what constant rate P, `.^(56)Mn` nuclei are being produced in the cyclontron during the bombardment?A. `2xx10^(11)"nuclei"//s1`B. `13.86xx10^(10)"nuclei"//s`C. `9.6xx10^(10)"nuclei"//s`D. `6.93xx10^(10)"nuclei"//s` |
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Answer» Correct Answer - B In equilibrium, rate of decay=Rate of production |
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| 255. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `^(56)Mn` is `.^(56)Mn` +`d` rarr `.^(56)Mn` +`p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `l n2=0.693`, Avagadro number` =6 xx 10^(23)`, atomic weight of `.^(56)Mn`=`56 g mol^(-1))`. After the activity of `.^(56)Mn` becomes constant, number of `.^(56)Mn` nuclei present in the target is equal to .A. `5xx10^(11)`B. `20xx10^(11)`C. `1.2xx10^(14)`D. `1.8xx10^(15)` |
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Answer» Correct Answer - D As Rate of decay=Rate of production `P= lambda N rArr N=(P)/(lambda)=(Pt_(1//2))/(ln 2)= 1.8xx10^(15)` |
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| 256. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`). After a long time bombardment, number `.^(56)Mn` nuclei present in the target depends upon.A. All (i), (ii) and (iii) are correctB. Only (i) and (ii) are correctC. Only (ii) and (iii) are correctD. Only (i) and (iii) are correct. |
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Answer» Correct Answer - C As `N=(Pt_(1//2))/(ln 2)`. It is dependent upon P and `t_(1//2)`. In equilibrium, rate of production=rate of decay. Large initial number will only make equilibrium come sonner. Hence, Initial number of `.^(56)Mn` nuclei will make no difference |
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| 257. |
In a thermo nuclear reaction `10^(-3)Kg` of hydrogen is converted into `0.99xx10^(-3)Kg` of helium. If the efficiency of the generator is `50%`, the electrical energy generated in `KWH` isA. `10^(5)`B. `1.5xx10^(5)`C. `1.25xx10^(5)`D. `1.3xx10^(5)` |
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Answer» Correct Answer - C `(E)/(t)=(mc^(2))/(t) , "effieiency"=("output")/("input")` |
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| 258. |
In nuclear fusion, One gram hydrogen is converted into `0.993 gm`. If the efficiency of the generator be `5%`, then energy obtained in `KWH` isA. `8.75xx10^(3)`B. `4.75xx10^(3)`C. `5.75xx10^(3)`D. `3.73xx10^(3)` |
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Answer» Correct Answer - A Efficency `= ("output")/(Deltamc^(2))` |
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| 259. |
Thermonuclear fusion reactions may becomes the source of untimited power for the mankind. A single fusion event involving isotopes f hydrogen produces more energy then energy from nuclear fission of `._93^235 U`.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. decreases with mass number at high mass numbersC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - C ( c) When fusion is achieved by raising the temperature of a system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour, it is called thermonuclear fusion. It is clean source of energy released in one fusion is much less than a sigle uranium fission. |
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| 260. |
Element `._(z)M^(A)` emits one `alpha` (alpha) particle followed by two `beta`( beta) particles. Among the following the daughter element isA. `._(z-2)M^(A-4)`B. `._(z-2)M^(A)`C. `._(z)M^(A-4)`D. `._(z+2)M^(a-4)` |
| Answer» Correct Answer - C | |
| 261. |
The half-life of a sample of a radioactive substance is `1` hour. If `8 xx 10^10` atoms are present at `t = 0`, then the number of atoms decayed in the duration `t = 2` hour to `t = 4` hour will beA. `2xx10^(10)`B. `1.5xx10^(10)`C. ZeroD. Infinity |
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Answer» Correct Answer - B |
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| 262. |
What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of useable energy ?. |
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Answer» `235` am u of uranium gives `185 MeV `energy. Therefore, the energy given by `1` am u of `._(92)U^(235)` is `(185)/(2.35) MeV=1(85)/(235) xx1.6 xx10^(-13) J` But `1` am u `=1.66 xx 10^(-27) kg`. Therefore, energy released by `1.66 xx 10^(-27)kg` of`._(92)U^(235)` is `(185 xx 1.6 xx 10^(-13)/(235)` Hence, energy released by `2 kg` of `._(92)U^(235)` , `W=(185 xx 1.6 xx10^(-13) xx2)/(235 xx1.66 xx10^(-27)) =1.517 xx 10^(14) J` Therefore , Therefore , pwer output of reactor `=(W)/(t) =(1.517 xx 10^(14))/((30 days))` `=(1.517 xx1`0^(14))/(30 xx 24 xx 50 xx 60)` `=5. 785 xx 10^(7) W`. |
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| 263. |
What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of usable energy ?.A. 5.585 MWB. 58.5 KWC. .585MWD. None of these |
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Answer» Correct Answer - A |
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| 264. |
Which one of the following unclear reactions is a source of energy in the sun ?A. `._4^9 Be + ._2^4 He rarr ._6^12 C + ._0^-1 n`B. `._2^3 He + ._2^3 He rarr ._2^4 He ._1^1 H + ._1^1H`C. `._56^144 Ba + ._56^92 Kr rarr ._92^235 U + ._0^-1 n`D. `._26^56 Fe + ._48^112 Ca rarr ._74^167 W + ._0^-1 n` |
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Answer» Correct Answer - B (b) Energy is released in the sun due to fusion. |
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| 265. |
Suppose we consider a large number of continers each containing initially `10000` atoms of a radioactive material with a half life of `1` year. After `1` year.A. All the containers will have `5000` atoms of the materialB. all the container will contain the same number of atoms of the material but that number will only be approximately `5000`C. the containers will in general have different numbers of the atoms of the material but their average will be close to `5000`D. none of the containers can have more than `5000` atoms. |
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Answer» Correct Answer - C Since radioactivity is a probability dependant phenomenon, nothing can be said decisicely. |
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| 266. |
Consider two arbitaray decay equation and mark the correct alternative` s` given below. (i) `._(92)^(230)U rarr n+._(92)^(229)U` (ii)`._(92_^(230)U rarr P +._(91)^(229)Pa` Given: `M(._(92)^(230)U) =230.033927 u,M(._(92)^(229)U)=229.03349 u, m_(n)=1.008665u`, `M(._(91)^(229)Pa) =229.032089, m_p`=1.007825, 1 am u =931.5 MeV`.A. Only decay (i) is possible.B. Only decay (ii) is possible.C. Both decay are possible.D. Neither of the two decay is possible. |
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Answer» Correct Answer - d For decay (i): `Q=[230.033927 -229.033496 - 1.008665]xx931.5` `= -7.7 MeV` (ii): `Q=[230.033927 - 229.032089 -1.007825] xx 931.5` `= -5.6 MeV` As `Q` is negative for both the decays, so none of the decay is allowed. |
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| 267. |
Let `E_(1)` and `E_(2)` be the binding energies of two nuclei `A` and `B`. it is observed that nuclei of `A` combine together to form a `B` nuclus. This observation is correct only if .A. `E_(1) gt E_(2)`B. `E_(2) gt E_(1)`C. `E_(2) gt 2E_(1)`D. nothing can be said |
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Answer» Correct Answer - c Transformation occurs only when the same net energy is released, which is possible only when `E_(2) gt2E_(1).` |
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| 268. |
In a smaple of rock, the ration `.^(206)Pb` to `.^(238)U` nulei is found to be `0.5.` The age of the rock is (given half-life of `U^(238)` is `4.5 xx 10^(9)` years).A. `2.25 xx 10^(9)year`B. `4.5 x 10^(9)` ln 3 yearC. `4.5 x 10^(9) (ln(3)/(2))/(ln2) year`D. `2.25 xx 10^(9) ln((3)/(2))` year |
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Answer» Correct Answer - c Suppose an initial radionuclide I decays to a final product F with a half-life `T_(1//2)` At any time, `N_(1) =N_(0) e^(- lambda `t`)` Number of product nuclei `=N_(F) =N_(0) -N_1` `(N_(F))/(N_(I))=(N_(0)- N_(I))/(N_I)=((N_(0))/(N_(I) -1))` `(N_(0))/(N_(I))=(1 +(N_(F))/(N_(I)))=1 +0.5=1.5` `e^(- lambda t)=1.5 rArr lambda t=1n 1.5` `:. (T_(1//2) In(1.5))/(1n 2) =4.5 xx 10^(9) (In ((3)/(2)))/(In2) year`. |
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| 269. |
The number of `U^(238)` nuclei in a rock sample equal to the number of `Pb^(206)` atoms. The half life of `U^(238)` is `4.5xx10^(9)` years. The age of the rock isA. `4.5xx10^(9)y`B. `9xx10^(9) y`C. `13.5xx10^(9)y`D. `18xx10^(9)y` |
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Answer» Correct Answer - A Given `N=(N_(o))/(2) , :. (N)/(N_(0))=(1)/(2)` that means `t=1` half life |
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| 270. |
In the uranium ore, the ratio of `U^(238)` nuclei to `Pb^(206)` nuclei is `2.8`. If it is assumed that all the lead `Pb^(206)` to be a final decay product of the uranuium series, the age of the ore is `[T_(1//2)` for `U^(238)` is `4.5xx10^(9)` years]A. `4.5xx10^(9)` yearsB. `2.0xx10^(9)` yearsC. `3.2xx10^(9)` yearsD. `6.4xx10^(9)` years |
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Answer» Correct Answer - B `U^(238)rarrPb^(206)` (radio nuclide) (stable nuclide) number of `Pb^(206)` nuclei=number of decayed uranium nuclei=`N_(0)(1-e^(-lambda t))` In present sample `(N(U^(238)))/(N(Pb^(206)))=2.8 rArr(N_(0)e^(-lambda t))/(N_(0)(1-e^(-lambda t)))=(28)/(10)` solve for time, t. |
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| 271. |
The number of `^238 U` atoms in an ancient rock equals the number of `^206 Pb` atoms. The half-life of decay of `^238 U` is `4.5 xx 10^9 y`. Estimate the age of the rock assuming that all the `^206 Pb` atoms are formed from the decay of `^238 U`. |
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Answer» Correct Answer - `[4.5 xx 10^(9)y` old] |
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| 272. |
In an ore containing Uranium, the ratio of `U^(238)` to `Pb^(206` nuceli is `3`. Calculate the age of the ore, assuming that alll the lead present in the ore is the final stable, product of `U^(238)`. Take the half-like of `U^(238)` to be `4.5 xx 10^(9)` years. In `(4//3) = 0.288`. |
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Answer» Correct Answer - `[1.867 xx 10^(9) "years"]` |
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| 273. |
An element` X` decays , first by positron emission and then two `alpha`-particles are emitted in successive radiactive decay. If the product nucleus has a mass number `229` and atomic number `89`, the mass number and atomic number of element ` X` are.A. `237,93`B. `237,94`C. `221,84`D. `237,92` |
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Answer» Correct Answer - b `._X^AXoverset("Proton")(rarr)._(Z-1)^AYoverset(2alpha)rarr._(A-5)^(A-8)Y` Given: `A-8 =224` and ` Z-5 =89` `rArr A=237,Z=94`. |
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| 274. |
The ratio isotope `._(29)^(64)Cu` can decay through positron emission and electron capture. In both the processes neutrino (v) is emitted. Calculate the difference in maximum possible kinetic energy of emitted neutrinos in the two processess. |
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Answer» Correct Answer - `(k_("max"))_(Ec)-(k_("max"))_(beta^(+))=2m_(e)c^(2)=1.02MeV` |
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| 275. |
Nuclei of `.^(64)Cu` can decay be electron capture (probability 61%) or by `beta^(+)` decay (39%). Half life of `.^(64)Cu` is 12.7 hour. Find the partial half life for electron capture decay oricess, |
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Answer» Correct Answer - 20.8hr. |
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| 276. |
During mean life of a radioactive element, the fraction that disintegrates isA. eB. `1/e`C. `(e-1)/e `D. `e/(e-1)` |
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Answer» Correct Answer - C |
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| 277. |
At any instant, the ratio of the amounts of two radioactive substance is `2:1`. If their half-lives be, respectively, `12h` and `16h`, then after two days, what will be the ratio of the substances?A. 0.042361111111111B. 0.084027777777778C. 0.043055555555556D. 0.044444444444444 |
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Answer» Correct Answer - A |
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| 278. |
If a `H_2` nucleus is completely converted into energy, the energy produced will be around.A. 1 MeVB. 938 MeVC. 9.38 MeVD. 238 MeV |
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Answer» Correct Answer - B (b) Mass of `H_2` nucleus = mass of proton `= 1 `amu energy equivalent to 1 amu is `931 MeV` so correct option is (b). |
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| 279. |
Which sample `A` or `B` shown in figure has shorter mean-life ? A. `tau_(B) lt tau_(A)`B. `tau_(B) gt tau_(A)`C. `tau_(B) = tau_(A)`D. Nothing can be concluded |
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Answer» Correct Answer - A at `t=0` `|(dN)/(dt)|_(A)=|(dN)/(dt)|_(B)` but `|(d^(2)N)/(dt^(2))|_(B) gt |(d^(2)N)/(dt^(2))|_(A)` `rArr lambda_(B) gt lambda_(A) rArrtau_(A) lt tau_(A)` |
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| 280. |
`He_(2)^(3)` and `He_(1)^(3)` nuclei have the same mass number. Do they have the same binding energy ?A. YesB. No, `._(2)^(3)He` has greater binding energy than, `._(1)^(3)He` due to extra proton.C. No, `._(2)^(3)He` has lower binding energy than, `._(1)^(3)He` due to extra proton.D. Cannot the concluded |
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Answer» Correct Answer - C Proton-Proton force, 12. `(dN)/(dt)= -lambda N` |
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| 281. |
Radon, `._(86)^(222)Rn`, is a radioactive gas that can be trapped in the basement of homes, and its presence in high concnetrations is a known health hazard. Radon has a half-life of `3.83 days`. A gas sample contains `4.0 xx 10^(8)` radon atoms initially. (a) How many atoms will remain after `12 days` have passed if no more radon leaks in? (b) What is the initial activity of the radon sample? |
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Answer» (a) A more precise answer is obtained by first finding the decay cosntant from equation `lambda =(0.693)/(T_(1//2))=(0.693)/(3.83 days)=0.181 days^(-1)` Taking `N_(0) =4 xx10^(8) `and the value of `lambda` just found to obtain the number N remaining after 12 days as `N=N_(0)e^(-lambdat)=-(4.0xx10^(8) "atoms") e^(-(0.181 days-1)(12 daus))` `=4.6 xx10^(7)"atoms"` This is vey close to our original estimate of `5.0 xx10^(7) "atoms"`. (b) First, we must express the decay constant in units of `s^(-1)` Using this value of `lambda`, we find that the initial activity as `R= lambda N_(0)=(2.09 xx10^(-6) s^(-1))(4.0 xx 10^(8))` `=840 decays s^(-1) =840 Bq`. |
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| 282. |
Ten grams of `^57 Co` kept in an open container beta-decays with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearlyA. `10 g`B. `7.5 g`C. `5 g`D. `2.5 g` |
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Answer» Correct Answer - A The weight will not change appreciable as the process is `beta`-decay. |
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| 283. |
Two isotopes of radioactive Radon gas `-R_(n)^(222)` and `R_(n)^(220)` are mixed in atomic ratio 1000:1 and kept in a `831cm^(3)` container at `10^(5) N//m^(2)` pressure and `27^(@)C` temperature the two isotopes have half life of 4 days and 1 minute respectively. Calculate the activity of Radon gas sample in unit of curie. |
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Answer» Correct Answer - `7.3xx10^(6)Ci` |
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| 284. |
Four different radioactive elements are kept in separated containers. In the beginning the container `A` has `200 g`-atom with half-life of `2` days, `B` has `20 g`-atom with half-life of `20` days, `C` has `2g`-atom with half-life `200` days and `D` has `100g`-atoms wityh half-life of `10` days. In the begining the maximum activity exhibited by the container isA. `A`B. `B`C. `C`D. `D` |
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Answer» Correct Answer - A The activity of radioactive element is given by `R= lambdaN=(0.631N)/(T)`Thus, `R prop (N)/(T)`. The value of `(N)/(T)` is largest for the element in the container `A`. |
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| 285. |
Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, `_(1)^(2)H` , known as deuteron and denoted by `D`, can be thought of as a candidate for fusion rector . The `D-D` reaction is `_(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+` energy. In the core of fusion reactor, a gas of heavy hydrogen of `_(1)^(2) H` is fully ionized into deuteron nuclei and electrons. This collection of `_1^2H` nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time `t_(0)` before the particles fly away from the core. If `n` is the density (number volume ) of deuterons , the product` nt_(0) `is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then `5 xx 10^(14) s//cm^(2)` it may be helpfull to use the following boltzmann constant `lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm` In the core of nucleus fusion reactor , the gas become plasma because ofA. strong nuclear force acting between the deutronsB. Coulomb force acting between the deutronsC. Coulomb force acting between deuteron-electron pairsD. the high pairs temperature maintained inside the reactor core |
| Answer» Correct Answer - D | |
| 286. |
When a boron nucleus (`_5^10B`) is bombarded by a neutron, an `alpha`-particle is emitted. Which nucleus will be formed as a result?A. `._6 C^12`B. `._3 Li^6`C. `._3 L i^7`D. `._4 Be^9` |
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Answer» Correct Answer - C ( c) `._5 B^10 + ._0 n^1 rarr ._3 Li^7 + ._2 He^4`. |
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| 287. |
As compared to `^12C` atom, `^14C` atoms hasA. Two extra protons and two extra electronsB. Two extra protons but no extra electronsC. Two extra neutrons and no extra neutrons and no extra electrons and two extra electronsD. Two extra neutrons and no extra electrons |
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Answer» Correct Answer - C ( c) For `._6 C^12, p = 6, e = 6, n = 6` For `._6 C^14, p = 6, e = 6, n = 8`. |
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| 288. |
`U^(238)` is found to be in secular equilibrium with `Ra^(226)` in its ore. If chemical analysis shown `1` atoms of `Ra^(226)` per `2.8xx10^(6)` atoms of `U^(238)`, find the half-like of `U^(228)`. Given the half life `Ra^(226)` is `1620` years. |
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Answer» For secular equilibrium `lambda_(U)N_(U) = lambda_(Ra)N_(Ra)` `(ln2)/((T_(1//2))_(U))N_(U) = (ln2)/((T_(1//2))_(Ra))N_(Ra)` `(2.8xx10^(6))/((T_(1//2))_(U)) = (1)/(1620)` `(T_(1//2))_(U) = 4.5 xx 10^(9)` year |
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| 289. |
Three `alpha-`particle and one `beta-`particle decaying takes place in series from an isotope `._88 Ra^238`. Finally the isotope obtained will be.A. `._(84)X^(220)`B. `._(86)X^(222)`C. `._(83)X^(224)`D. `._(83)X^(215)` |
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Answer» Correct Answer - C |
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| 290. |
`4` grams of radioactive substance `A` left `1//2 gm` after some ture. `1` gran of another radioactive substance `B` left `1//4 gm` in the same period, If half life of `B` is `2` hours, the half life of `A` is ( in hours)A. `3//4`B. `4//3`C. `1//4`D. `1//2` |
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Answer» Correct Answer - B `W=(W_(0))/(2^(n))`, no of half lives `n=(t)/(t_(1//2))` |
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| 291. |
One mole of `alpha` emitter of half life equal to `2` days was placed in a sealed tube for `4` days at `S.T.P` volume of helium collector isA. `22.4 lit`B. `16.8 lit`C. `11.2 lit`D. `5.6 lit` |
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Answer» Correct Answer - B `W=(W_(0))/(2^(n)), n (t)/(t_(1//2))`. initially value of Helium `22.4` lits no of half lives `=2` |
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| 292. |
Which of the following is a correct statement ?A. Beta rays are same as cathode raysB. Gamma rays are high energy neutronsC. Alpha particle are singly ionized helium atomsD. Protons and neutrons have exactly the same mass. |
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Answer» Correct Answer - A (a) Both the `beta-`rays and the cathode rays are made up of electrons. `gamma-`rays are `EM` waves, `alpha-`particles are doubly ionized helium atoms and protons and neutrons have approximately the same mass. |
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| 293. |
After `1 alpha` and `2 beta` emissions.A. Mass number reduces by `3`B. Mass number reduces by `4`C. Mass number reduces by `6`D. Atomic number remains unchanged. |
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Answer» Correct Answer - B (b) `._Z X^A overset (alpha)rarr ._(Z -2) Y^(A -4) overset (2 beta)rarr ._Z X^(A -4)`. |
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| 294. |
After `1 alpha` and `2 beta` emissions.A. mass number reduces by `2`B. mass number reduces by `6`C. atomic number reduces by `2`D. atomic number remains unchanged |
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Answer» Correct Answer - D |
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| 295. |
Alpha particles areA. `2` free protonsB. helium atomsC. singly ionised helium atomsD. doubly ionised helium atoms |
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Answer» Correct Answer - D |
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| 296. |
Complete the equation for the following fission process `._92 U^235 ._0 n^1 rarr ._38 Sr^90 +....`.A. `._(54)Xe^(143)+3 ._(0)n^(1)`B. `._(54)Xe^(145)`C. `._(57)Xe^(142)`D. `._(54)Xe^(142)+ ._(0)n^(1)` |
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Answer» Correct Answer - A |
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| 297. |
Write the decay equations and expression for the disintegration energy` Q` of the following decay: `beta-` decay,electron capture. |
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Answer» In all the cases discussed here, we neglect any neutrino mass. `beta-` decay: `M_(nucl) (._(Z)^(X) X) =M_(nucl) (._(Z+1)^(A) D) + m_(e) +Q//c^(2)` Where `M_(nucl)` indicates the nuclear mass. In order to convert it to atomic mass, we add `Zm_(e)` on both sided: `M._(nucl) (._Z^(AX)+Zm._(e) =M._(nucl) (._(Z+1)^(A)D)+(Z+1) m._(e)+Q//c^(2)` Since atomic binding energies are less than nuclear binding energies, they are neglected on the two sides of the equation. `[M((._Z)^(A)X)=M(._(Z+1)^(A)D)]c^(2)` `Q` for this equation `=[M(._Z^AX)-M(._(Z+1)^(A)D)]c^(2)` `beta+` decay: `M_(nucl) (._Z^AX)=M_(nucl) (._(Z-1)^(A)D) +m._(e) +Q//c^(2)` In this case, only `Z-1 m` is needed for the daughter atomic mass which gives us a remaining mass of `2 m_(e)` `[M(._Z)^(A)X)-M (._(Z-1)^(A)D)-2m_(e)]c^(2)` `Q` for this equation` =[M(._Z)^AX)-M (._(Z-1)^AD)-2m_(e)]c^(2)` We add `(Z-1) m_(e)` to each side above and obtain `M._(nucl)(._Z^AX)-M (._(Z-1)^AD)+Q//c^(2)` `Q` for this equtation`=[M(._Z)^AX)-M (._(Z-1)^AD)-2m_(e)]c^(2)` Electron capture`. |
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| 298. |
Find whether `alpha-` deacy or any of the `beta-` decay are allowed for `._(89)^(226)Ac`. |
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Answer» Our first step will be to write the reaction, then find the disintegration energy `Q`. If Q lr 0, the decay is allowed. `alpha-` decay: `(._(89)^(226)) Ac rArr .(_(87)^(222)Fr) + alpha` `Q=[M(._89)^(226)Ac)-M (._(87)^(222)Fr)-M.^(4He)]c^(2)` `=5.50 MeV` `beta-` deacy: `._(89)^(226) Ac) rarr (._(90)^(226)Th )+beta- + bar(V)` `Q=[M(._89)^(226)Ac)-M (._(90)^(226)Th)]c^(2)` `1.12 MeV` `beta+` decay: `(._(89)^(226)AC)rarr._(88)^(226)Ra) +beta+ +v` `Q=[M(._89)^(226)Ac)-M (._(88)^(226)Ra)-2m_(e))]c^(2)` `=-0.38 MeV` Electron capture : `(._(89)^(226)Ac) + e rarr ._(88)^(226)Ra +v` `Q=[M(._89)^(226)Ac)-M (._(88)^(226)Ra)]c^(2)` `=0.64 MeV` (Electron capture is allowed). |
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| 299. |
A radioactive nucleus `A` finally transforms into a stable nuccelus `B`. Then, `A` and `B` may beA. isobarsB. isotonesC. isotopesD. none of these |
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Answer» Correct Answer - c A and `B` can be isotopes if number of `beta`decays is two times the number of `alpha`-decays. |
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| 300. |
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. `1`B. `2`C. `4`D. All of the above |
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Answer» Correct Answer - a Let `lambda_(A) =lambda and lambda_(B) =2 lambda`. Initially, rate of disintergaration of `A` is `lambda N_(0)` and that of `B` is `2 almbda N_(0)`. After one half-life of `A`, rate of disntigration of `A` will come become `lambdaN_(0)//2` (half-life of `B =one` half the half -life of `A`. So, after one half-life of A or two half-lives of `B`, `((- dN)/(dt))=((-dN)/(dt))_(B)` `:. n=1`. |
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