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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.
| 551. |
8g of the radioactive isotope, cesium -137 were collected on February 1 and kept in a sealed tube. On July 1, it was found that only 0.25 g of it remained. So the half-life period of the isotope isA. 37.5 daysB. 30 daysC. 23 daysD. 50 days |
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Answer» Correct Answer - B t = Feb 1 to July 1 `= 28 + 31 + 30 + 31 + 30 = 150` days `lamda = (2.303)/(150) "log" (8)/(0.25) = (2.303)/(150) log 2^(5) = (0.693)/(30) "day"^(-1)` `t_(1//2) = (0.693)/(0.693//30)` = 30 days. |
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| 552. |
The half-life period `t_(1//2)` of a radioactive element is N years. The period of its complete decays isA. `N^(2)` yearsB. 2N yearsC. `(1)/(2)N^(2)` yearsD. Infinity |
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Answer» Correct Answer - D The `t_(1//2)` of a radioactive element = N years `:.` The period of its complete decay is infinity |
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| 553. |
A radioactive nuclide X decays at the rate of `1.00 xx 10^(5)` disintegration `S^(-1) g^(-1)`. Radium decays at the rate of `3.70 xx 10^(10)` disintegration `s^(-1) g^(-1)`. The activity of X in millicurie `g^(-1) (mci g^(-1))` isA. 0.027B. `0.270 xx 10^(-5)`C. 0.0027D. 0.00027 |
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Answer» Correct Answer - B `(dx_(1))/(dt) = lamda N_(1), 1 xx 10^(5) = lamda N_(1)` `(dx_(2))/(dt) = lamda N_(2), 3.7 xx 10^(10) = lamda N_(2)` `(N_(1))/(N_(2)) = (1 xx 10^(5))/(3.7 xx 10^(10)) = (1 xx 10^(-5))/(3.7) = 0.27 xx 10^(-5)` |
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| 554. |
A radioisotope has half life of 10 years. What percentage of the original amount of it would you expect to remain after 20 years? |
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Answer» Correct Answer - C `t_(1//2) = 10` years, `T = 20` year We know `T = nt_(1//2) implies n = (20)/(10) = 2` And `N = ((1)/(2))^(n) N_(0)` `(N)/(N_(0)) = ((1)/(2))^(2) = (1)/(4)` % of `(N)/(N_(0)) = (1)/(4) xx 100 = 25%` |
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| 555. |
A radioactive isotope decays at such a rate that after 96 minutes only `(1)/(8)th` of the original amount remains. The half-life of this nuclide in minutes isA. 12B. 24C. 32D. 48 |
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Answer» Correct Answer - C `lamda = (2.303)/(t) "log" ([N_(0)])/([N]) = (2.303)/(96) "log" (1)/(1//8)` `:. T_(1//2) = (0.693)/(lamda) = (0.693)/(0.0216) = 32.0` min |
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| 556. |
A radioisotope has a half life of 10 days. If tofay there is `125 g ` of it left, what was its mass 40 days earlier ?A. `600g`B. `1000g`C. `1250g`D. `2000g` |
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Answer» Correct Answer - D `T=nxxt_(1//2)` `:. n=(40)/(10)=4` `N=((1)/(2))^(n)N_(0)` `:. N_(0)=125xx(2)^(4)=125xx16=2000` |
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| 557. |
A radioisotope has `t_(1//2) = 3` years. After a given amount decays for 12 years, what fraction of the original isotope remains? |
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Answer» Given half life `(t_(1//2)) = 3` years Time for decay `(T) = 12` years We know that `T = n xx t_(1//2)` `12 = n xx 3` `:.n = (12)/(3) = 4` Let the original amount `N_(0)` Let the amount left after three half-life periods be `N`. we know that `N = ((1)/(2))^(n) N_(0)` or `(N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(4) = (1)/(16)` |
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| 558. |
The disintegration constant of radium with half-life 1600 years isA. `2.12 xx 10^(-4)"year"^(-1)`B. `4.33 xx 10^(-4)"years"^(-1)`C. `3.26 xx 10^(-3)"year"^(-1)`D. `4.33 xx 10^(-12)"year"^(-1)` |
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Answer» Correct Answer - B `k = (0.693)/(t_(1//2)) = (0.693)/(1600) = 4.33 xx 10^(-4) "year"^(-1)` |
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| 559. |
`Ra^(226)` has half life of 1600 years. The number of disintegration per second per gram isA. `3.7xx10^(10)`B. `9.2xx10^(6)`C. `3.7xx10^(9)`D. `3.7 xx 10^(8)` |
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Answer» Correct Answer - A `K=(Numbe rof d i s i nt e g rations per se c o nd)/(To tal n umber of at oms is on e gram of Ra)` `K=(0.693)/(t_(1//2))` or `K=(0.693)/(1600xx365xx24xx60xx60)` `:. ` Number of disintegration per second `=(0.693xx6.023xx10^(23))/(1600xx365xx24xx60xx60xx226)` `=3.7xx10^(10)` |
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| 560. |
Radium has atomic weight 226 and a half-life of 1600 Yr. The number of disintegrations produced per second from one gram areA. `4.8xx10^(10)`B. `9.2xx10^(6)`C. `3.7xx10^(10)`D. zero |
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Answer» Correct Answer - c Given, M = 226, `t_(1//2) = 1600 yrs, w = 1 g` `(-dN)/(dt) = lambda(6.023xx10^(23)xxw)/(M) = (0.693)/(t_(1//2))xx(6.023xx10^(23)xx1)/(226)` `= (0.692xx6.023xx10^(23))/(1600xx226xx(365xx24xx3600))` `3.7xx10^(10)` disintegrations per second |
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| 561. |
`(15)/(16)th` of a radioactive sample decays in 40 days. Half-life of the sample isA. 100 daysB. 10 daysC. 1 dayD. `log_(e) 2` days |
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Answer» Correct Answer - B Quantity of radioactive element decayed = `(15)/(16)` Quantity left `= 1 - (15)/(16) = (1)/(16)` `(1)/(16) = 1 xx ((1)/(2))^(n) or ((1)/(2))^(4) = ((1)/(2))^(n)` one half-life `= (40)/(4) = 10` days. |
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| 562. |
A radioactive sample decays to half of its initial concetration in 6.93 minutes. It further decays half in next 6.93 minutes. The rate constant for the reaction isA. `0.10 "min"^(-1)`B. `0.01 "min"^(-1)`C. `1.0 "min"^(-1)`D. `0.001 "min"^(-1)` |
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Answer» Correct Answer - A `k = (0.693)/(t_(1//2)) = (0.693)/(6.93) = 0.10 "min"^(-1)` |
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| 563. |
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. The binding energy per nucleon of `._(1)H^(2)` and `._(2)He^(4)` are `1.1 MeV` and `7 MeV`, respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released isA. `13.9 MeV`B. `26.9 MeV`C. `23.6 MeV`D. `19.3 MeV` |
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Answer» Correct Answer - C `E = Delta m xx 931 MeV`, where `Delta m =` mass defect |
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| 564. |
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 MeV`B. `3.27 MeV`C. `0.82 MeV`D. `2.45 MeV` |
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Answer» Correct Answer - B `E = Delta m xx 931 MeV` |
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| 565. |
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. Mass equivalent to the energy `931 MeV` isA. `6.02 xx 10^(-27) kg`B. `1.662 xx 10^(-27) kg`C. `16.66 xx 10^(-27) kg`D. `16.02 xx 10^(-27) kg` |
| Answer» Correct Answer - B | |
| 566. |
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. Fusion reaction takes place at aboutA. `9 xx 10^(2) K`B. `3 xx 10^(3) K`C. `3 xx 10^(4) K`D. `3 xx 10^(6) K` |
| Answer» Correct Answer - D | |
| 567. |
The radioactively due to `C^(14)` isotope (half-life 6000 years) of a sample of wood from an ancient tomb was found to be nearly half that of fresh wood, the tomb is therefore aboutA. 3000 years oldB. 6000 years oldC. 9000 years oldD. 1200 years old |
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Answer» Correct Answer - B `N = N_(0) xx ((1)/(2))^(n)` `(1)/(2) = 1 xx ((1)/(2))^(n) , n = 1` `t = n xxt_(1//2) = 1 xx 6000 = 6000` yrs |
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| 568. |
1.0 g of a radioactive isotope was found to reduce to 125 mg after 24 hours. The half-life of the isotope isA. 8 hoursB. 24 hoursC. 6 hoursD. 4 hours |
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Answer» Correct Answer - A `N = [(1)/(2)]^(n) xx N_(0) = 125mg = ((1)/(2))^(n) xx 1000 mg` `((1)/(2))^(n) = (125)/(1000) = (1)/(8)` `((1)/(2))^(n) = ((1)/(2))^(3), n = 3`, so number, of `t_(1//2) = 3` Total time = 24 hours, Half-life time = `(24)/(3)` = 8 hours. |
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| 569. |
The radioactivity due to `C-14` isotope (half-life = 6000 years) of a sample of wood form an ancient tomb was found to be nearly half that of fresh wood. The bomb is there for aboutA. 3000 year oldB. 6000 year oldC. 9000 year oldD. 12000 year old |
| Answer» Correct Answer - B | |
| 570. |
A radioactive element has half life of `4.5 xx 10^(9)` years. If `80 g` of this was taken, the time taken for it to decay to `40 g` will be a. `2.25 xx 10^(9)` years b. `4.50 xx 10^(9)` years c. `6.75 xx 10^(9)` years d. `8.75 xx 10^(9)` yearsA. a. `2.25 xx 10^(9)` yearsB. b. `4.50 xx 10^(9)` yearsC. c. `6.75 xx 10^(9)` yearsD. d. `8.75 xx 10^(9)` years |
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Answer» Correct Answer - b. `4.50 xx 10^(9)` years `t_(1//2) = 4.5 xx 10^(9)` years Here the amount decay to 50% i.e., from `80 g` to `40 g` Therefore time required will be `4.5 xx 10^(9)` years. |
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| 571. |
The half-life of `.^(14)C` is aboutA. 12.3 yearsB. 5730 yearsC. `4.5 xx 10^(9)` yearsD. `2.52 xx 10^(5)` years |
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Answer» Correct Answer - B `t_(1//2) " of " C^(14) = 5730` years |
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| 572. |
The decay constant of a radioactive element is `3 xx 10^(-6) "min"^(-1)`. Its half-life isA. `2.31 xx 10^(5)` minB. `2.31 xx 10^(6)` minC. `2.31 xx 10^(-6)` minD. `2.31 xx 10^(-7)` min |
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Answer» Correct Answer - A `t_(1//2) = (0.693)/(lamda) = (0.693)/(3 xx 10^(-6) "min"^(-1)) = 2.31 xx 10^(5)` min |
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| 573. |
What mass of `Ra^(226)` whose `t_(1//2) = 1620` years will give the activity of 1 millicurie? |
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Answer» Activity `= (0.692)/(t_(1//2)) xx (W)/("Atomic weight") xx 6.023 xx 10^(23)` `3.7 xx 10^(7) = (0.693)/(226) xx 6.023 xx 10^(23)` or `W = 10^(-3) g` |
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| 574. |
If equal numer of atoms of two radioactive elements are considered, the most dangerous would be the one with a half life of? a. 4.0 million years b. 100 years c. 0.01 second d. 1 secondA. a. 4.0 million yearsB. b. 100 yearsC. c. 0.01 secondD. d. 1 second |
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Answer» Correct Answer - c. 0.01 second Lesser the half of isotope, less stable will be the iostope. |
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| 575. |
The half-life of radium (226) is 1620 years. The time taken to convert 10 grams of radium to 1.25 grams isA. 810 yearsB. 1620 yearsC. 3240 yearsD. 4860 years |
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Answer» Correct Answer - D `t = (2.303 xx t_(1//2))/(0.693) "log" (N_(0))/(N)` |
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| 576. |
The half life of radium (226) is 1620 years. The time takend to convert `10 g` of radium to `1.25 g` is a. 810 years b. 1620 years c. 3240 years d. 4860 yearsA. a. 810 yearsB. b. 1620 yearsC. c. 3240 yearsD. d. 4860 years |
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Answer» Correct Answer - d. 4860 years `t_(1//2) = 1620` years `N_(0) = 1.25 g` We know `K = (2.303)/(t) log (N_(0))/(N)` `implies (0.693)/(t_(1//2)) = (2.303)/(t) log (10)/(1.25)` `:.t = (2.303 xx 1620)/(0.693) log (10)/(log (10)/(1.25)) = (0.3)/(log 8) = (0.3)/(3 xx 0.3)` `:. t_(x%) = t_(1//2) xx 3 = 1620 xx 3 = 4860` years |
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| 577. |
After 20 min, the amount of certai radioactive substance disintegrate was 15/16 original amount. What is the half-life ofg the radioactive substance? |
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Answer» Correct Answer - 5 `2""^(n)=N""_(0)/N,n=4,n=t/t""_(1//2)` |
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| 578. |
Which of the following Statement `is//are` not true?A. Among radioactive rays, `alpha-`rays has the highest penetrating powerB. Penetrating power of `beta-`rays is greater than `gamma-`raysC. `alpha-`rays is not deflected in magnetic fieldD. `alpha-` rays has highest momentum among radioactive rays |
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Answer» Correct Answer - `(A,B,C)` |
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| 579. |
Fluorine- 18 is one of the radioactive nuclides utilised in Positron Emission Tomography (PET) scans. Fluorine-18 can be synthesized by bombarding oxygen-18 nuclei with hydrogen-1 nuclei, `"Oxygen"-18 + "Hydrogen" -1 to ` `"Fluorine"- 18+X` What is the identity of the other product, `X,` in this reaction?A. Alpha particleB. Beta particleC. PositronD. Neutron |
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Answer» Correct Answer - D |
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| 580. |
Correct set of four quantum numbers for the valence (outermost) electron of rubidium `(Z = 37)` isA. `n=5, l=0, m=0, s=+1//2`B. `n=5, l=1, m=0, s=+1//2`C. `n=5, l=1, m=1, s=+1//2`D. `n=6, l=0, m=0, s=+1//2` |
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Answer» Correct Answer - A I : for `n=5, l_(min)=0, :.` Orbital angular momentum `=sqrt(l(+1)) ħ=0`. (False) II : Outermost electronic configuration `=3s^(1)` or `3s^(2)`. :. possible atomic number `=11` or `12` (False). III : `Mn_(25)=[Ar] 3d^(5) 4s^(2) :. 5` unpaired electrons. :. Total spin `=+- 5/2` (False). IV : Inert gases have no unpaired electrons. :. Spin magnetic moment `=0` (True). |
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| 581. |
A piece of wood was found to have `C^(14)//C^(12)` ratio 0.6 times that in a living plant. Calculate that in a living plant. Calculate the period when the plant died. (Half life of `C^(14) = 5760` years)? |
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Answer» We know that `t = (2.303 xx t_(1//2))/(0.693) "log" (N_(0))/(N)` So `t = (2.303 xx 5760)/(0.693) "log" (1)/(0.6)` `= (2.303 xx 5760)/(0.693) xx 0.2201` `= 4213 "years"` |
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| 582. |
A piece of wood was found to have `C^(14)//C^(12)` ratio 0.7 times that in a living plant. The time period when the plant died is (Half-life of `C^(14)` = 5760 yr)A. 2770 yrB. 2966 yrC. 2980 yrD. 3070 yr |
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Answer» Correct Answer - B Given, `N_(0) = 1N, N_(t) = 0.70 and t_(1//2) = 57600 yr` `k = (0.693)/(t_(1//2)) = (0.693)(5760)` We also know, `k = (2.303)(t) "log" (N_(0))/(N_(t)) = (0.693)/(5760)` or `t = (2.303 xx 5760 xx 0.155)/(0.693) = 2966yr` |
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| 583. |
A radioactive element has half-life of one day. After three days, the amount of the element left will beA. 1.2 of the original amountB. 1.4 of the original amountC. 1.8 of the original amountD. 1/16 of the original amount |
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Answer» Correct Answer - C `n = (3)/(1) = 3, N = (N_(0))/(2^(3)) = (1)/(8)` |
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| 584. |
A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the roomA. 1000 daysB. 300 daysC. 10 daysD. 100 days |
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Answer» Correct Answer - D Activity `(-(dN)/(dt)) prop N` `N = N_(0) ((1)/(2))^(n) rArr (N)/(N_(0)) = ((1)/(2))^(n)` `(1)/(10) = ((1)/(2))^(n) rArr 10 = 2^(n)` `log 10 = n log 2 rArr n = (1)/(0.301) = 3.32` `t = n xx t_(1//2) = 3.32 xx 30 = 99.6` days |
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| 585. |
A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room ?A. `underline~`1000 daysB. `underline~`300 daysC. `underline~`10 daysD. `underline~`100 days |
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Answer» Correct Answer - d It will be safe to enter in the room when activity gets reduces by 10 times, i.e, when N=`N""_(0)//10` `therefore N""_(0)/N=((1)/(2))""^(n)or1/10=(1/2)""^(n)` `or10=2""^(n)` log10=nlog2 `n=1/0.301=3.32 t=nlog2` `=3.32xx30=99.6 days` |
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| 586. |
A sample of rock moon contains equal numbers of atoms of uranium and lead `t_(1//2)` for `U` is `4.5 xx 10^(9)` years. The age of rock would be a. `4.5 xx 10^(9)` years b. `9.0 xx 10^(9)` years c. `13.5 xx 10^(9)` years d. `2.25 xx 10^(9)` years |
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Answer» `t = (2.303)/(K) "log" [1 + (Pb^(206))/(U^(238))]` `implies (2.303 xx 4.5 xx 10^(9))/(0.693) "log" [1 + ((1)/(206))/((1)/(238))]` `:.t = 13.5 xx 10^(9)` years |
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| 587. |
A sample of rock from moon contains equal number of atoms of uranium and lead `(t""_(1//2)for U=4.5xx10""^(9)`year). The age of the rock would beA. `4.5xx10""^(9)` yearB. `9xx10""^(9)` yearC. `13.5xx10""^(9)` yearD. `2.25xx10""^(9)` year |
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Answer» Correct Answer - a `lamdat=2.303log((U""^(238)+pb""^(206))/(U""^(238)))` `lamda=0.693/t""_(1//2),txx0.693/t""_(1//2)=2.303 log2,t=t""_(1//2)` `N=N""_(0)/2""^(n),2""^(n)=N""_(0)/N=15.2/7.6=2,n=1` |
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| 588. |
The excitation energy of first excited state of a hydrogen like atom is `40.8 eV`. Find the energy needed to remove the electron to form the ion. |
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Answer» Correct Answer - `54.4 eV` `40.8=(Delta E)_(2 to 1)xxZ^(2) rArr 40.8=10.2xxZ^(2) rArr z^(2)=4 "or" Z=2` `IE=13.6 Z^(2)=13.6xx4=54.4 eV` |
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| 589. |
Light of wavelength `lambda` strikes a metal surface with intensity `X` and the metal emits `Y` electrons per second of average energy `Z`. What will happen to `Y` and `Z` if `X` is havled?A. T will be halvedB. Y will doubleC. Y will be remain sameD. Z will be halved |
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Answer» Correct Answer - 1 Number of emitted electron `prop` Intensity of incident light. |
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| 590. |
de Broglie proposed dual nature for electron by putting his famous equation `lambda = (h)/(mv).` Later on, Heisenberg proposed uncertainty principle as `deltapDeltax ge (h)/(4pi).` On the contrary, Particle nature of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface it gives up its energy to the electron. Part of this energy (say W) is used by the electrons to escape from the metal and the remaining energy imparts kinetic energy `(1//2 mv^(2))` to the ejected photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential . When a beam of photons of a particular energy was incident on a surface of a particular pure metal having work function =(40 eV), some emitted photoelectrons had stopping potential equal to 22 V. some had 12V and rest had lower values. Calculate the wavelength of incident photons assuming that at least one photoelectron is ejected with maximum possible kinetic energy :A. `310 Å`B. `298 Å`C. `238 Å`D. `200 Å` |
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Answer» Correct Answer - D The maximum `KE` of potoelectron is corresponding to maximum stopping `=22 eV` `:. E_("incident")=E_("threshold")+KE_(maxi)=40 eV+22 eV=62 eV` `lambda_("incident")=(12400 Å)/(62)=200 Å` |
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| 591. |
de Broglie proposed dual nature for electron by putting his famous equation `lambda = (h)/(mv).` Later on, Heisenberg proposed uncertainty principle as `deltapDeltax ge (h)/(4pi).` On the contrary, Particle nature of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface it gives up its energy to the electron. Part of this energy (say W) is used by the electrons to escape from the metal and the remaining energy imparts kinetic energy `(1//2 mv^(2))` to the ejected photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential . The circumference of third orbit of a single electron species is 3 nm. What may be the approximate wavelength of the photon required to just ionize electron from this orbit?A. `19.2xx10^(-2) m`B. `5.76xx10^(-2) m`C. `3.84xx10^(-2) m`D. `1.92xx10^(-2) m` |
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Answer» Correct Answer - D `Deltax=(h)/(4pi Me)xx1/(DeltaV)" "DeltaV=V 0.001/100=300xx10^(-5) m//s` `Deltax=5.8xx10^(-5)xx1/(300xx10^(-5))=1.92xx10^(-2) m` |
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| 592. |
If uranium `(` mass number 238 and atomic number 92`)` emits an `alpha-`paticle, the produc has mass number and atomic numberA. 236 and 92B. 234 and 90C. 238 and 90D. 236 and 90 |
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Answer» Correct Answer - B Emission of an `alpha-` particle means mass is decreased by 4 units and charge byy 2 units. Thus, `._(92)U^(238)overset(-alpha)rarr._(90)U^(234)` Thus, the mass number `=234` Atomic number `=90` |
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| 593. |
If uranium `(` mass number 238 and atomic number 92`)` emits an `alpha-`paticle, the produc has mass number and atomic numberA. `236 and 92`B. `234 and 90`C. `238 and 90`D. `236 and 90` |
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Answer» Correct Answer - B |
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| 594. |
In the disintegration of a radioactive element, `alpha`- and `beta`-particles are evolved from the nucleus. `._(0)n^(1) rarr ._(1)H^(1) + ._(-1)e^(0) +` Antineutrino + Energy `4 ._(1)H^(1) rarr ._(2)He^(4) + 2 ._(+1)e^(0) +` Energy Then, emission of these particles changes the nuclear configuration and results into a daughter nuclide. Emission of an `alpha`-particles results into a daughter element having atomic number lowered by 2 and mass number by 4, on the other hand, emission of a `beta`-particle yields an element having atomic number raised by 1. Select the correct statements among the following:A. Emission of a `beta`-particle results into an isobar of parent element.B. Emission of `alpha`-particles results into an isodiapher of parent element.C. Emission of one `alpha`-and two `beta`-particles results into an isotope of the parent element.D. Emission of `gamma`-radiations may yield a nuclear isomer. |
| Answer» Correct Answer - A::B::C::D | |
| 595. |
Assertion (A) : Nucleus of the atom does not contain electrons,yet it ewmits `beta`-particles in the form electrons Reason (R ) : In the nucleus, protons and neutrons exchange mesons frequentlyA. If both (A) and (R ) are correct, and (R ) is the correct explaination of (A)B. If both (A) and (R ) are correct, but (R ) is not the correct explaination of (A)C. If (A) is correct,but (R ) is incorrectD. If both (A) and (R ) are incorrect. |
| Answer» Correct Answer - c | |
| 596. |
Match the entries listed in Column I with appropriate entries listed in Column II |
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Answer» Correct Answer - `A rarr s; B rarr r; C rarr q; D rarr p` (A) `._(7)^(14)N + ._(2)^(4) x rarr ._(8)^(17)O + ._(1)^(1)p` (B) `._(1)^(2)D + ._(83)^(209)Bi rarr ._(84)^(210)Po + ._(0)^(1)x` (C) `._(1)^(1) p + _(4)^(9)Be rarr ._(4)^(8)Be + ._(1)^(2)x` (D) `._(0)^(1) n + ._(11)^(23)Na rarr ._(11)^(24)Na + ._(0)^(0)gamma` |
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| 597. |
The number of neutrons emitted when`._(92)^(235)U` undergoes controlled nuclear fission to `._(54)^(142)Xe and ._(38)^(90)Sr` is |
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Answer» Correct Answer - 3 `._(92)U^(235) rarr ._(54)Xe^(142) + ._(38)Sr^(90) + 3 ._(0)n^(1)` |
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| 598. |
If a `._(92)U^(235)` nucleus upon being struck by a neutron changes to `._(56)Ba^(145)`, three neutrons and an unknown product. What is the unknown product? |
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Answer» The reaction can be presented as: `._(92)U^(235) ._(0)n^(1) rarr ._(56)Ba^(145) + ._(Z)X^(A) + ._(0)n^(1)` Equating the mass number on both sides, we get `235 + 1 = 145 + A + 3 xx 1` `:. A = 88 =` Atomic mass of `X` Atom Similarly equating atomic numbers on both sides, we get `:. 92 + 0 = 56 + Z + 3 xx 0` `:. Z = 36 =` Atomic number of `X` Atoms Therefore, unknown product is `._(36)X^(88)`, i.e., `._(36)Kr^(88)`. |
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| 599. |
How many `alpha`-and `beta`-particle will be emitted when `._(98)Th^(234)` change into `._(84)Po^(218)`? |
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Answer» `underset("Parent")(._(90)Th^(234) rarr ._(84)Po^(218)` Decrease in mass `= (234 - 218) = 16` Mass of 1 `alpha-"particle" `= 4 amu Therefore, number of `alpha`-particles emitted Number of `beta`-emitted - (Atomic number `= 2 xx` Number of `beta`-particles - Atomic number of end product) `2 xx 4 - (90 - 84) (8 - 6) = 2` Hence, number of `alpha`-particles = 4 and number of `beta`-particles = 2 |
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| 600. |
`._(92)X^(234) (-7alpha)/(-6beta)Y.` Find out atomic number, mass number of `Y` and identify it. |
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Answer» Correct Answer - `._(84)PO^(206)` |
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