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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.
| 101. |
Binding energy per nucleus of `._(1)^(2)H` and `._(2)^(4)He` are 1.1 MeV and 7 MeV respectively. Calculate the amount of energy released in the following process: `._(1)^(2)H + ._(1)^(2)H to ._(2)^(4)He` |
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Answer» Amount of energy released `= sum` Binding energy of products `- sum` Binding energy of reactions `=[4 xx 7] - [4 xx 1.1] = 23.6 MeV` |
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| 102. |
How many `alpha` and `beta` particles should be eliminated so that an isodiaphere is formed?A. `n alpha, n beta`B. `n alpha, (n + 1) beta`C. `n alpha`D. `n beta` |
| Answer» Correct Answer - c | |
| 103. |
which of the following are used as control rods in a nuclear reacto ?A. Cadmium rodB. Graphite rodsC. steel rodsD. All of these |
| Answer» Correct Answer - a | |
| 104. |
Which of the following notation shows the product incorrectly?A. `._(96)Cm^(242) (alpha, 2n) ._(97)Bk^(243)`B. `._(5)B^(10) (alpha, n) ._(7)N^(13)`C. `._(7)N^(14) (n,p) ._(6)C^(14)`D. `._(14)Si^(28) (d, n) ._(15)P^(29)` |
| Answer» Correct Answer - a | |
| 105. |
Which is ture about decay constant `(lambda)`?A. Unit is `"time"^(-1)`B. value of `lambda` is always less than 1.C. `lambda` is independent of temperature.D. `lambda` is defined as the ratio of no. of atoms disintegrating per unit time to the total no. of atoms present at the time. |
| Answer» Correct Answer - ac | |
| 106. |
At radioactive equilibrium, the ratio between two atoms of radioactive elements `A` and `B` is `3.1 xx 10^(9) : 1`. If the half-life period of `A` is `2 xx 10^(10)` years, what is the half-life of `B`?A. 30 yrsB. 3 yrsC. 3.3 yrsD. None of these |
| Answer» Correct Answer - c | |
| 107. |
Which of the following is not correct?A. Nuclei of atoms participate in nuclear reactionsB. `._(20)^(40)Ca` and `._(18)^(40)Ar` are isotopesC. 1 amu of mass defect is approximately equal to 931.5 MeVD. Uranium `(U^(238))` series is knowns as `(4n + 2)` series. |
| Answer» Correct Answer - b | |
| 108. |
In the atmosphere, carbon dioxide is found in two forms, i.e., `.^(12)CO_(2)` and `.^(14)CO_(2)`. Plants absorb `CO_(2)` during photosynthesis. In presence of chlorophyll, plants synthesise glucose. `6 CO_(2) + 6 H_(2) O to overset(hv)(to) C_(6)H_(12)O_(6) + 6O_(2) uarr` Half life of `.^(14)C` is 5760 years. The analysis of wooden artifacts for `.^(14)C` and `.^(12)C` gives useful information for deermination of its age. all living organisms, because of their constant exchange of `CO_(2)` with the surrounding have the same ratio of `.^(14)C` to `.^(12)C`, i.e., `1.3 xx 10^(-12)`. When an organism dies, the `.^(14)C` in it keeps on decaying as follows: `._(6)^(14)C to ._(7)^(14)N + ._(-1)^(0)e` + Energy Thus, the ratio `.^(14)C//^(12)C` decrease with the passage of time. we can be used to date anything made of organic matter, e.g., bone, skeleton, wood, etc. Using carbon dating material have been dated to about 50,000 years with accuracy. A piece of wood from an archaeological source shows a `.^(14)C` activity which is 60% of the activity found in fresh wood today. The age of archaeological sample will be:A. 4246 yrsB. 4624 yrsC. 4628 yrsD. 6248 yrs |
| Answer» Correct Answer - a | |
| 109. |
In the atmosphere, carbon dioxide is found in two forms, i.e., `.^(12)CO_(2)` and `.^(14)CO_(2)`. Plants absorb `CO_(2)` during photosynthesis. In presence of chlorophyll, plants synthesise glucose. `6 CO_(2) + 6 H_(2) O to overset(hv)(to) C_(6)H_(12)O_(6) + 6O_(2) uarr` Half life of `.^(14)C` is 5760 years. The analysis of wooden artifacts for `.^(14)C` and `.^(12)C` gives useful information for deermination of its age. all living organisms, because of their constant exchange of `CO_(2)` with the surrounding have the same ratio of `.^(14)C` to `.^(12)C`, i.e., `1.3 xx 10^(-12)`. When an organism dies, the `.^(14)C` in it keeps on decaying as follows: `._(6)^(14)C to ._(7)^(14)N + ._(-1)^(0)e` + Energy Thus, the ratio `.^(14)C//^(12)C` decrease with the passage of time. we can be used to date anything made of organic matter, e.g., bone, skeleton, wood, etc. Using carbon dating material have been dated to about 50,000 years with accuracy. In the process of photosythesis, `O_(2)` gas is released form:A. `CO_(2)`B. `H_(2)O`C. both `H_(2)O` and `CO_(2)`D. mechanism is not confirmed |
| Answer» Correct Answer - b | |
| 110. |
In the atmosphere, carbon dioxide is found in two forms, i.e., `.^(12)CO_(2)` and `.^(14)CO_(2)`. Plants absorb `CO_(2)` during photosynthesis. In presence of chlorophyll, plants synthesise glucose. `6 CO_(2) + 6 H_(2) O to overset(hv)(to) C_(6)H_(12)O_(6) + 6O_(2) uarr` Half life of `.^(14)C` is 5760 years. The analysis of wooden artifacts for `.^(14)C` and `.^(12)C` gives useful information for deermination of its age. all living organisms, because of their constant exchange of `CO_(2)` with the surrounding have the same ratio of `.^(14)C` to `.^(12)C`, i.e., `1.3 xx 10^(-12)`. When an organism dies, the `.^(14)C` in it keeps on decaying as follows: `._(6)^(14)C to ._(7)^(14)N + ._(-1)^(0)e` + Energy Thus, the ratio `.^(14)C//^(12)C` decrease with the passage of time. we can be used to date anything made of organic matter, e.g., bone, skeleton, wood, etc. Using carbon dating material have been dated to about 50,000 years with accuracy. A wooden piece is 11520 yrs old. What is the fraction of `.^(14)C` activity left in the piece?A. 0.12B. 0.25C. 0.5D. 0.75 |
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Answer» Correct Answer - b `(0.693)/(t_(1//2) .^(14)C) = (2.303)/(t_("age")) log ((N_(0))/(N))` `(0.693)/(5760) = (2.303)/(11520) log ((N_(0))/(N))` `(N)/(N_(0)) = 0.25` |
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| 111. |
Nathan Thomson, one of the the first inhabitants of lord howe Island. Decided to plant some eruopean deciduous trees in his garden. Unifortunately the exact timing of planting the seeds is not known, over the years, pollen produced by the trees accumulated at the bottom of the lake near Nathan,s house. Very small quantities of radioactive `.^(210)Pb` (`t_(1//2)` = 22.3 years) were deposited at the same time. Note that european deciduous trees pollinate in their first year of growth. In 1995, a team of researchers sampled a sediment core from the bottom of the lake. the examination of sediment core of found that: (a) Pollen of trees first occurs at the depth of 50 cm. The activity of `.^(210)Pb` at the top of sediment core is 356 Bq/kg and at 50 cm depth 1.40 Bq/kg. In what year did Nathan Thomson plant the seeds?A. `1719 +- 2`B. `1819 +- 2`C. `1519 +- 2`D. `1919 +- 2` |
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Answer» Correct Answer - b `(0.693)/(t_(1//2)) = (2.303)/(t_("age")) log ((N_(0))/(N))` `(0.693)/(22.30) = (2.303)/(t_("age")) log ((356)/(1.40))` `t_("age") = 176` yrs |
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| 112. |
Thiosulphate ion `(S_(2)O_(3)^(2-))` on acidification changes to `SO_(2)` along the precipitation of sulpur, `.^(35)S^(32) SO_(3)^(2-) + 2H^(+) to H_(2)O + SO_(2) + S` Which is the correct statement?A. `S^(35)` is in sulphurB. `S^(35)` is in `SO_(2)`C. `S^(35)` is in bothD. `S^(35)` is in none |
| Answer» Correct Answer - a | |
| 113. |
An analysis of the rock shows that the relative number of `Sr^(87)` and `Rb^(87) (t_(1//2)=4.7 xx10^(10)` year) atoms is 0.05 . What is the age of the rock? Assume all the `Sr^(87)` have been formed from `Rb^(87)` only |
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Answer» `(.^(87)Rb)/(.^(87)Sr) = (x)/(y) = (1)/(0.052)` `(x)/(x + y) = (1)/(1.052)` `(N)/(N_(0)) = (1)/(1.052)` `lambda = (2.303)/(t) log_(10) (N_(0))/(N)` `(0.693)/(4.7 xx 10^(10)) = (2.303)/(t) = (2.303)/(t) log_(10) 1.052` `t = 3.43 xx 10^(9)` years |
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| 114. |
Which of the following nuclei are doubly magic?a)`._(92)U^(238)` b)`._(2)He^(4)`c)`._(8)O^(16)`d)`._(82)Pb^(208)`A. `._(2)^(4)He`B. `._(8)^(16)O`C. `._(82)^(208)Pb`D. `._(92)^(238)U` |
| Answer» Correct Answer - a,b,c | |
| 115. |
The activity of radioisotope changes with:A. temperatureB. pressureC. chemical enviormentD. none of these |
| Answer» Correct Answer - d | |
| 116. |
The half life of `._(92)^(238)U` is `4.5 xx 10^(9)` years. Uranium emits an `alpha`particle to give thorium. Calculate the time required to get the product which contains equal masses of thorium and uranium. |
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Answer» `N_(0) = (1)/(238) + (1)/(234)`, `N = (1)/(238)` Use, `(0.693)/(t_(1//2)) = (2.303)/(t) log_(10) ((N_(0))/(N))` |
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| 117. |
32 mg of pure `._(94)^(238)PuO_(2)` has an activity of `6.4 xx 10^(7) sec^(-1)` (i) What will be the half life of `._(94)^(238)Pu` in years? (ii) What amount of `PuO_(2)` will remain if 100 mg `PuO_(2)` is kept for 500 years? |
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Answer» (i) Mass of `.^(238)Pu = (238)/(270) xx 32 = 28.207 mg` Rate `= (0.693)/(t_(1//)) xx (w)/("At.wt") xx N` `6.4 xx 10^(7) = (0.693)/(t_(1//2)) xx (28.207 xx 10^(-3))/(238) xx 6.023 xx 10^(23)` `t_(1//2) = 7.729 xx 10^(11) sec = 2.45 xx 10^(4)` years (ii), `(0.693)/(t_(1//2)) = (2.303)/(t) log_(10) ((N_(0))/(N))` `(0.693)/(2.45 xx 10^(4)) = (2.303)/(5000) log_(10) ((100)/(N))` `N = 86.7` mg |
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| 118. |
On analysis a sample of uranium are was found to contain 227g of `._(82)Pb^(208)` and 1.667 g of `._(92)U^(238)`. The half life peiod of `U^(238)` is `4.51 xx 10^(9)` yrs. If all the lead was assumed to have come from decay of `._(92)U^(238)`, what is the age of the earth? |
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Answer» Mole of `U^(238) = (1.667)/(238)` Moles of `Pb^(206) = (0.277)/(206)` `N_(0) = (1.667)/(238) + (0.277)/(206)` `N = (1.667)/(238)` `t = (2.303)/(lambda) log (N_(0))/(N)` `= (2.303 xx 4.51 xx 10^(9))/(0.693) log_(10) ((1.667)/(238) + (0.277)/(206))/((1.667)/(238))` `= 1.143 xx 10^(9)` years |
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| 119. |
A sample of uranium mineral was found to contain `Pb^(208)` and `U^(238)` in the ratio of 0.008 : 1. Estimate the age of the mineral (half life of `U^(238)` is `4.51 xx 10^(9)` years). |
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Answer» We know that, `t = (2.303t_(1//2))/(0.693) log [1 + (.^(206)Pb)/(.^(238)U)]` Given `t_(1//2) = 4.51 xx 10^(9)` years Ratio by mass of `.^(206)Pb : .^(238)U = 0.008 : 1` Ratio by moles of `.^(206)Pb : .^(238)U = (0.008)/(206) : (1)/(238) = 0.0092` So, `t = (2.303 xx 4.51 xx 10^(9))/(0.693) log [1 + 0.0092]` `= (2.303 xx 4.51 xx 10^(9))/(0.693) xx 0.00397` `= (0.0412)/(0.693) xx 10^(9) = 0.05945 xx 10^(9)` years Hence, age of the mineral is `5.945 xx 10^(7)` years. |
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| 120. |
The half-life period of `U^(234)` is `2.5 xx 10^(5)` years. In how much time is the quantity of the isotope reduce to 25% of the original amount? |
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Answer» Initial amount of this isotope `N_(0) = 100` Final amount of the isotope N = 25 We know that, `N = ((1)/(2))^(n) N_(0)` So, `25 = ((1)/(2))^(n) xx 100` or `(25)/(100) = ((1)/(2))^(n)` or `(1)/(4) = ((1)/(2))^(n)` or `((1)/(2))^(2) = ((1)/(2))^(n)` or n = 2 Time taken `T = n xx t_(1//2)` `= 2 xx 2.5 xx 10^(5) = 5 xx 10^(5)` years |
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| 121. |
The `.^(60)C` isotope decays with a half life of 5.3 years. How long would it take for 7/8 of a sample of 500 mg of `.^(60)Co` to disintegrate?A. 21.2 yearsB. 15.9 yearsC. 10.6 yearsD. 5.3 years |
| Answer» Correct Answer - b | |
| 122. |
A radioisotope has a half life of 10 days. If today there is `125 g ` of it left, what was its mass 40 days earlier ?a)600gb)1000gc)1250gd)2000gA. 600gB. 1000gC. 1250gD. 2000g |
| Answer» Correct Answer - d | |
| 123. |
There are four radioactive decay series called thorium `(4n)`, uranium `(4n + 2)` actinium `(4n + 3)` and neptunium `(4n + 1)` series. Neptunium series is artificial while other three series are natural. The end productsd of each radioacitve decay series have stable nuclei. All natural decay series terminate at lead but neptunium or artificial series terminate at bismuth. Actinium series begins with an isotope ofa)Actiniumb)Radiumc)Uraniumd)PoloniumA. actiniumB. radiumC. uraniumD. polonium |
| Answer» Correct Answer - c | |
| 124. |
The disintegration constant of `.^(238)U` is `1.54 xx 10^(-10) "years"^(-1)`. Calculate the half life period of `.^(238)U`. |
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Answer» Half life period `t_(1//2) = (0.693)/(lambda)` since, `lambda = 1.54 xx 10^(-10) "years"^(-1)` So, `t_(1//2) = (0.693)/(1.54 xx 10^(-10)) = 4.5 xx 10^(9)` years |
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| 125. |
A radioisotope has `t_(1//2) = 5` years. After a given amount decays for 15 years, what fraction of the original isotope remains? |
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Answer» Half life `(t_(1//2) = 5` years Time for decay `(T) = 15` years We know that, `T = n xx t_(1//2)` So, `15 = n xx 5` or n = 3 We know that, `N = ((1)/(2))^(n) N_(0)` or `(N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(3) = (1)/(8)` Thus, after 15 years `(1)/(8)th` of the original amount remains. |
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| 126. |
Starting from radium, the radioactive disintegration process terminates when the following is obtaineda)radonb)leadc)uraniumd)thoriumA. leadB. radonC. radium AD. radium B |
| Answer» Correct Answer - a | |
| 127. |
Radium has a half life 1600 years and its daughter elements radon has a half life 3.82 days. In an enclosure, the volume of radon was found constant for a week. Explain and calculate the ratio of the number of radium and radon nuclei. Will the ratio be constant after 400 years? |
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Answer» `(N_(1)(Ra))/(N_(2)(Rn)) = (t_(1//2)(Ra))/(t_(1//2)(Rn))` `(N_(1))/(N_(2)) = (1600)/(3.82) xx 365` `= 1.528 xx 10^(5)` |
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| 128. |
The half-life period of radium is 1600 years. Calculate the disintegration of radium. |
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Answer» Disintegration constant `lambda = (0.639)/(t_(1//2))` since, `t_(1//2) = 1600` years So, `lambda = (0.693)/(1600)` or `lambda = 4.33 xx 10^(-4) "years"^(-1)` |
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| 129. |
Radium has atomic weight 226 and a half-life of 1600 Yr. The number of disintegrations produced per second from one gram areA. `4.8 xx 10^(10)`B. `3.7 xx 10^(8)`C. `9.2 xx 10^(6)`D. `3.7 xx 10^(10)` |
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Answer» Correct Answer - d `("No. of disintegration per sec")/("Total no. of atoms in one gram of Ra")` `= (0693)/(1600 xx 365 xx 24 xx 60 xx 60)` or No. of disintegration per sec `= (0.693 xx 6.032 xx 10^(23))/(1600 xx 365 xx 24 xx 60 xx 60 xx 226)` |
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| 130. |
The source of energy of stars is nuclear fusion. Fusion reaction occurs at very high temperature, about `10^(7) `. Energy released in the process of fusion is due to mass defect. It is also called `Q`-value. `Q = Delta mc^(2), Delta m =` mass defect. In a nuclear reaction `._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1)` If the masses of ._(1)H^(2)` and `._(2)He^(3)` are 2.014741 and 3.016977 amu, respectively. then the `Q`-value of the reaction is nearly.A. 0.00352 MeVB. 3.27 MeVC. 0.82 MeVD. 2.45 MeV |
| Answer» Correct Answer - b | |
| 131. |
The isotope `._(92)^(235)U` decays in a number of steps to an isotope of `._(82)^(207)Pb`. The groups of particle emitted in this process will be:A. `4 alpha, 7 beta`B. `6 alpha, 4 beta`C. `7 alpha, 4 beta`D. `10 alpha, 8 beta` |
| Answer» Correct Answer - c | |
| 132. |
The end product of `4n` series isa)`._(82)Pb^(208)`b)`._(82)Pb^(207)`c)`._(82)Pb^(209)`d)`._(82)Pb^(204)`A. `._(82)^(208)Pb`B. `._(82)^(207)Pb`C. `._(82)^(209)Pb`D. `._(83)^(210)Bi` |
| Answer» Correct Answer - a | |
| 133. |
The correct starting material and product of different disintegration series `"is"//"are"`a)`Th^(232), Pb^(208)`b)`Np^(237),Bi^(209)`c)`U^(235),Pb^(206)`d)`U^(238),Pb^(206)`A. `.^(232)Th, .^(208)Pb`B. `.^(235)U, .^(206)Pb`C. `.^(238)U, .^(207)Pb`D. `.^(237)Np, .^(209)Bi` |
| Answer» Correct Answer - a,d | |
| 134. |
`._(90)^(232)Th` disintegrates to `._(82)^(208)Pb`. How many of `beta`-particle are evolved? |
| Answer» Correct Answer - 4 | |
| 135. |
The decay constant for an `alpha-` decay of `Th^(232)` is `1.58xx10^(-10)s^(-1)`. How many `alpha-`decays occur from `1 g` sample in 365 days ?A. `2.89 xx 10^(19)`B. `1.298 xx 10^(19)`C. `2.219 xx 10^(19)`D. None of these |
| Answer» Correct Answer - b | |
| 136. |
In `alpha-` decay , `n//p` ratio `:`a)May inrease or decreaseb) Remains constantc)Decreasesd)IncreasesA. may increase off decreaseB. remains constantC. decreasesD. increases |
| Answer» Correct Answer - d | |
| 137. |
In `beta-` decay `n//p` ratio` :`a)May inrease or decreaseb) Remains constantc)Decreasesd)IncreasesA. remains unchangedB. may increase or decreaseC. increasesD. decreases |
| Answer» Correct Answer - d | |
| 138. |
`beta`-particle is emitted in radioactivity byA. conversion of proton to neutronB. from outermost orbitC. conversion of neutron to protonD. `beta`-particle is not emitted |
| Answer» Correct Answer - c | |
| 139. |
Correct order of radioactivity is:A. `._(1)H^(1) gt ._(1)H^(2) gt ._(1)H^(3)`B. `._(1)H^(3) gt ._(1)H^(2) gt ._(1)H^(1)`C. `._(1)H^(3) gt ._(1)He^(1) gt ._(1)H^(2)`D. `._(1)H^(3) gt ._(1)H^(1) gt ._(1)H^(2)` |
| Answer» Correct Answer - b | |
| 140. |
Which of the following are correct with respect to the unit of radioactivity? (i) The SI unit of radioactivity is curie (Ci) (ii) `1 Ci = 3.7 xx 10^(-10)` dis `s^(-1)` (iii) 1 Bq `= 3.7 xx 10^(-10) Ci` (iv) The SI unit of radioactivity is becquerel (Bq) (v) 1 Ci ` = 3.7 xx 10^(10)` BqA. (i) and (iii)B. (iv) and (v)C. (i) and (ii)D. (ii) and (iv) |
| Answer» Correct Answer - b | |
| 141. |
Radio carbon dating is done by estimating in the specimen:A. the amount of ordinary carbon still presentB. the amount of radio carbon still presentC. the ratio of amount of `._(6)^(14)C` to `._(6)^(12)C` still presentD. The ratio of amount of `._(6)^(12)C` to `._(6)^(14)C` still present |
| Answer» Correct Answer - c | |
| 142. |
A plot of the number of neutrons `(n)` against the number of protons `(p)` of stable nuclei exhibits upward deviation from linearity for atomic number, `Zgt 20.` For an unstable nucleus having `n//p` ratio less than 1, the possible mode `(s)` of decay is `(are)`A. `beta^(-)`-decay (`beta` emission)B. orbital of K-electron captureC. neutron emissionD. `beta^(+)`-decay (positron emission) |
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Answer» Correct Answer - b,d When N/P ratio is lower than required for stabiltiy such nuclei can increase N/P ratio by adopting any one of the following mode of decay: (i) Positron `(beta^(+))` emission (ii) K-electron capture (iii) Emission of `alpha`-particles `:.` (b) and (d) options are correct. |
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| 143. |
Neutrons are more effective projectiles than protons because theyA. are attracted by nuclieB. are not repelled by nucleiC. travel with high speedD. none of these |
| Answer» Correct Answer - b | |
| 144. |
(A) Protons are better projectiles than neutrons. (R ) The neutrons being neutral do not experience repulsion from positively charged nucleus.A. If both (A) and (R ) are correct and (R ) is the correct explanation for (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A )C. If both (A) and (R ) are incorrect.D. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - D | |
| 145. |
(A) The nucleus of gold is stable even though there is a very strong coulombic repulsion among the protons. (R ) The inverse square coulomb force is exactly balanced by another inverse square force which is very powerful i.e., nuclear forceA. If both (A) and (R ) are correct and (R ) is the correct explanation for (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A )C. If both (A) and (R ) are incorrect.D. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - C | |
| 146. |
Assertion `(A):` `._(92)U^(238) (IIIB)overset(-alpha)rarrAoverset(-alpha)rarrB overset(-beta)rarrC` Reason `(R):` Element `B` will be of `II A ` group.a)If both `(A)` and `(R)` are correct , and `(R)` is the correct explanation of `(A)`b)If both `(A)` and `(R)` are correct, but (R) is not the correct explanation of `(A)`c)If `(A)` is correct, but `(R)` is incorrect.A. If both (A) and (R ) are correct and (R ) is the correct explanation for (A).B. If both (A) and (R ) are correct but (R ) is not the correct explanation for (A )C. If both (A) and (R ) are incorrect.D. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - B | |
| 147. |
`._(92)U^(238)` emits `8 alpha-` particles and `6 beta-` particles. The `n//p` ratio in the product nucleus isa)`(62)/(41)`b)`(60)/(41)`c)`(61)/(42)`d)`(62)/(42)`A. 60/41B. 61/40C. 62/41D. 61/42 |
| Answer» Correct Answer - c | |
| 148. |
Radioactive disintegration of `._(88)Ra^(226)` `Ra overset(-alpha)rarr Rn overset(-alpha)rarr RnA overset(-alpha)rarr RaB overset(-beta)rarr RaC` Determine the mass number, atomic number, and group in periodic table for `RaC`. |
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Answer» Parent element is `._(88)^(226)Ra`. atomic mass = 226 Atomic number = 88 RaC is formed after the emission of 3 alpha particles. Mass of 3 alpha particles `= 3 xx 4 = 12` So, Atomic mass of RaC = (226 - 12) = 214 With emission of one `alpha`particle atomic number is decreased by 2 and with the emission of one `beta`-particle, atomic number is increased by 1. So, Atomic number of RaC = `88 - (3 xx 2) + 1 = 83` |
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| 149. |
`._(92)^(238)Th` is a natural an `alpha`-emitter. After `alpha` emission, the residual `U_(X_(1))` in turns emits a `beta`-particles to produce nucleus `U_(X_(2))`. Find out the atomic number and mass number of `U_(X_(1))` and `U_(X_(2))`. Also if uranium belongs to IIIrd group to which group `U_(X_(1))` and `U_(X_(2))` belong. |
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Answer» `._(92)^(238)U - ._(2)^(4)He to ._(90)^(234)U_(X_(1))` `._(90)^(234)U_(X_(1)) - (._(-1)^(0)e) to ._(91)^(234)U_(X_(2))` Both `U_(X_(1))` and `U_(X_(2))` will belong to IIIrd group because both lie in actinide series. |
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| 150. |
In the transformation of `""_(92)""^(238)U` to `""_(92)""^(234)U`, if one emission is an `alpha` particle, what should be the other emission(s)?a)Two `beta""^(-)`Two `beta""^(-)`b)Two `beta""^(-)` and one `beta""^(-)`c)One `beta""^(-)` and one `gamma`d)One `beta""^(+)` and one `beta""^(-)`A. two `beta^(-)`B. two `beta^(-)` and one `beta^(+)`C. one `beta^(-)` and one `gamma`D. one `beta^(+)` and one `beta^(-)` |
| Answer» Correct Answer - a | |