InterviewSolution
Saved Bookmarks
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.
| 1. |
Calculate the equivalent weight of each oxidant and reducant in: (a) `FeSO_(4) + KCl_(3) rarr KCl + Fe_(2) (SO_(4))_(3)` (b) `Na_(2) SO_(3) + Na_(2)CrO_(4) rarr Na_(2) CrO_(4) rarr Na_(2) SO_(4) + Cr(OH)_(3)` (c) `Fe_(3)O_(4) + KMnO_(4) rarr Fe_(2) O_(3) + MnO_(2)` (d) `KI + K_(2) Cr_(2) O_(7) rarr Cr^(3+) + 3I_(2)` (e) `Mn^(4+) rarr Mn^(2+)` (f) `NO_(3)^(-) rarr N_(2)` (g) `N_(2) rarr NH_(3)` (h) `Na_(2) S_(2) O_(3) + I_(2) rarr Na_(2) S_(4) O_(6) + 2NaI` (i) `FeC_(2)O_(4) rarr Fe^(3+) + CO_(2)` |
|
Answer» Correct Answer - (a) `152,20.42`, (b) `63.54`, (c) `232,52.67`, (d) `166,49`, (e) `27.5`, (f) `12.4`, (g) `4.67`, (h) `158.127`, (i) `48` |
|
| 2. |
In the reaction `Al + Fe_(3)O_(4) rarr Al_(2) O_(3) + Fe` (a) Which element is oxidized and which is reduced? (b) Total no. of elecetros transferred during the change. |
|
Answer» Correct Answer - `Al` is oxidized and `Fe^(8//3+)` is reduced, (b) `24` |
|
| 3. |
If `m_(p)` and `m_(n)` are masses of proton and neutron respectively and `M_(1)` and `M_(2)` are masses of `._(10)MeNe^(20)` and `._(20)Ca^(40)` nucleus repectively, then:A. `M_(2) = 2M_(1)`B. `M_(2) lt 2M_(1)`C. `M_(2) gt 2M_(1)`D. `M_(1) lt 10(m_(n) + m_(p))` |
|
Answer» Correct Answer - `(b,d)` Binding energy of `(10p + 10n)` is lesser than `(20p + 20n)` |
|
| 4. |
The intergrated rate equation is `Rt = log, C_(0) - log C_(t)`. The straight line graph is obtained by plotting:A. `t` vs log `C_(t)`B. `(1)/(t)` vs log `C_(t)`C. `t` vs `C_(t)`D. `(1)/(t)` vs log `(1)/(C_(t))` |
|
Answer» Correct Answer - `(a)` `log C_(1) vst (log C_(t) = log C_(0) - Rt)` having a `y = c + mx` slope equal to `-R` |
|
| 5. |
The half life of `.^(215)At` is `100mu s`. The time taken for the radioacticity of `.^(215)At` to decay to `1//16^(th)` of its initial value is:A. `400mu s`B. `6.3mu s`C. `40mu s`D. `300mu s` |
|
Answer» Correct Answer - `(a)` `(1)/(16) = (1)/(2^(4)), :. n = 4, t_(1//2) = 100 mu s` `:. T = t_(1//2) xx n` `= 100xx4 = 400 mu s` |
|
| 6. |
The half life period of `._(53)I^(125)` is 60 days. What % of radioactivity would be present after 240 days. |
|
Answer» Correct Answer - `6.25 %` No. of half life`=(240)/(60)=4=("Given time")/(t_(1//2))` `N_(t)=(N_(0))/(2^(n))=(N_(0))/(2^(4))` `100xx(N_(t))/(N_(0))=(1)/(16)xx100rArr 6.25%` |
|
| 7. |
The disntergration rate of a certain radioactive sample at any instant is `4750 dp m`. Five minute later, the rate becomes `2700 dp m` Calculate half-line of sample. |
|
Answer» `r_(0) = 4750` dpm at `t = 0` `r_(t) = 2700 dpm` at `t = 5` min. `:. (r_(0))/(r_(t)) = (4750)/(2700)` Also Rate `prop` NO. of atoms `:. (r_(0))/(r_(t)) = (N_(o))/(N_(t)) = (4750)/(2700)` `:. T = (2.303)/(lambda) log_(10) (N_(0))/(N_(t))` `5 = (2.303)/(lambda) log_(10) (4750)/(2700)` `lambda = 0.113 "mintute"^(-1)` `:. t_(1//2) = (0.693)/(0.113) = 6.13` minute |
|
| 8. |
Two elements `P` and `Q` have half-line of `10` and `15` minutes repectively. Freshly preapared sample of mixuture containing equal number of atoms is allowed to decay for `30` minutes. The ratio of number of atoms of `P` and `Q` in left in mixture is:A. `0.5`B. `2.0`C. `3.0`D. `4.0` |
|
Answer» Correct Answer - `(a)` `N_(p) = N_(OP).e^((0.693xx30)/(10)) = N_(op).e^(-0.693xx3)` `N_(Q) = N_(oQ).e^(-(0.693xx30)/(15)) = N_(OQ).e^(-0.693xx2)` `:. (N_(p))/(N_(Q)) = e^(-0.693) = 0.5` |
|
| 9. |
`KMnO_(4)` oxidises `NO_(2)^(-)` to `NO_(3)^(-)` in basic medium. How many moles of `NO_(2)^(-)` are oxidised by `1 mol` of `KMnO_(4)`? |
|
Answer» Correct Answer - `1.5 mole NO_(2)^(-)` |
|
| 10. |
A freshly prepared radioactive source to half-life `2 hr`, emits radiatiosn of intensity which is 64 times the permissible safe level. The minumum time after which it would be possibleto work safely with the source is:A. `6 hr`B. `12 hr`C. `24 hr`D. `128 hr` |
|
Answer» Correct Answer - `(b)` `N = (N_(0))/(64) = (N_(0))/(2^(6)), :. N = 6` Thus total time `= 2xx6 = 12hr`. |
|
| 11. |
Detemine the oxidation no. of the folliwing elements given in bold letters: (a) `KmnO_(4)`, (b) `H_(2)SO_(5)` (c) `H_(2)S_(2)O_(8)`m (d) `NH_(4)NO_(3)` (e) `K_(4)Fe(CN)_(6)`, (f) `OsO_(4)` (g) `HCN`, (h) `HNC` (i) `HNO_(3)`, (j) `KO_(2)` (k) `Fe_(3)O_(4)`, (l) `KI_(3)` (m) `N_(3)H`, (n) `Fe(CO)_(5)` (o) `Fe_(0.94)O`, (p) `NH_(2) NH_(2)` (q) `FeSO.(4).(NH_(4))_(2)SO_(4) 6H_(2)O` (r) `Na_(2)[Fe(CN)_(5)NO]`, (u) `[Fe(NO)(H_(2)O)_(5)]SO_(4)` (v) `Na_(2)S_(4)O_(6)` (v) `Na_(2)S_(2)O_(3)` (w) Dimethyl sulphoxide or `(CH_(3))_(2) SO` (x) `Na_(2)S_(2)O_(3)`, (y) `CrO_(5)` or `CrO(O_(2))_(2)` (z) `CaOCl_(2)` |
|
Answer» Correct Answer - (a) `+7`, (b) `+6`, (c) `+6`, (d) `-3,+5`, (e) `+2`, (f) `+8`, (g) `+2,-3`, (h) `+2`, (i) `+5`, (j) `-(1)/(2)`, (k) `+(8)/(3)` , (l) `-(1)/(3)`, (m) `-(1)/(3)`, (n) `0`, (o) `(200)/(94)`, (p) `-2`, (q) `+2`, (r) `+3`, (s) `+7`, (t) `+2`, (u) `+1`, (v) `+(5)/(2)`, (w) `0`, (x) `+2`, (y) `+6`, (z) `-1,+1` |
|
| 12. |
The nuclei of two radioactive isotopes of same substance `A^(236)` and `B^(234)` are present in the ratio of `4:1` in an ore obtained form some other planet. Their half-lives are 30 and 60 minutes respectibvely. Both isotopes are alpha emitters and acitity of isotope `A^(236)` si 1 rutherford `(10^(6) dps)`. Calcualte: (a) After how much time their activities will becomes identical? (b) The time required to bring the ratio of their atoms to `1:1` |
|
Answer» Correct Answer - (a) `180` min., (b) `120` min |
|
| 13. |
The activity of the radioactive sample drops to 1/64 of its original value in 2 hr find the decay constant `(lambda)`. |
|
Answer» Correct Answer - `lambda=2.079hr^(-1)` `lambda=(1)/(2) ln. (1)/((1//64))=2.079 hr^(-1)` |
|
| 14. |
The activity of a radioactive sample decreases to `1//3` of the original activity, `A_(0)` in a period of 9 years. After 9 years more, its activity `A_(0)//x`. Find the value of `x`. |
|
Answer» Correct Answer - `9` `t = (2.303)/(lambda) log (A_(0))/(A_(1)0` `9 = (2.303)/(lambda) log 3 implies lambda = 0.122` `18 = (2.303)/(0.122) log (A_(0))/(A_(2))` `(A_(2))/(A_(0)) = (1)/(9), :. A_(2) = (A_(0))/(9) ("given" (A_(0))/(x))` `:. X = 9` |
|
| 15. |
Consider the following nuclear reactions : `._(92)^(238)M rarr _(Y)^(X)N+2 . _(2)^(4)He, ._(Y)^(X)N rarr _(B)^(A)L+2beta^(+)` The number of neutrons in the element L isA. 142B. 144C. 140D. 146 |
|
Answer» Correct Answer - B `._(92)^(238)M rarr _(Y)^(X)N+2._(2)^(4)He` Applying mass number balance & nuclear charge balance 92=y+4 `" "` y=88 238=x+8 ` " "` x=230 `._(Y)^(X)N rarr _(B) ^(A) L+2 beta^(+) (._(+1)e^(0))` Applying mass number balance & nuclear charge balance 230=A+0 `" "` A=230 88=B+2 ` " " ` B=86 No. of neutrons in element L is 230-86=144. |
|
| 16. |
Consider the follwing nuclear reactions: `._(92)^(238) M rarr ._(y)^(x)N + 2 ._(2)He^(4)` `._(y)^(x)N rarr ._(B)^(A)L + 2beta^(+)`A. `146`B. `144`C. `140`D. `142` |
|
Answer» Correct Answer - `(b)` `._(92)^(238)M rarr ._(88)^(230)N + 2 ._(2)He^(4)` `._(88)^(230)N rarr ._(86)^(230)L + 2 ._(+1)e^(0)` |
|
| 17. |
Rutherford studied the first nuclear reaction `[._(7)^(14)N(alpha,p) ._(8)^(17)O]` which take place with a change in energy equivalent to `1.193 MeV`. Later on various types of nuclear reactions such as artifical radioactivity, artifical transmutatoon , nuclear fission, nuclear fussion, spallation reactions etc. were studied. Nuclear fusion reactions is not:A. Uncontrolled reactionB. used in formation of `H` bombC. Thermo nuclear reactionsD. Carried out at low temperature |
|
Answer» Correct Answer - `d` It is a fact. |
|
| 18. |
Rutherford studied the first nuclear reaction `[._(7)^(14)N(alpha,p) ._(8)^(17)O]` which take place with a change in energy equivalent to `1.193 MeV`. Later on various types of nuclear reactions such as artifical radioactivity, artifical transmutatoon , nuclear fission, nuclear fussion, spallation reactions etc. were studied. The reaction `._(97)^(14)N + ._(2)^(4)He + ._(8)^(17)O + ._(1)^(1)H` may be carried out by bombaring `N` atoms with `alpha-` particles of energy:A. `= 1.193 MeV`B. `gt 1.193 MeV`C. `lt 1.93 MeV`D. `le 1.93 MeV` |
|
Answer» Correct Answer - `b` `alpha-` particle need more energy to penetrate nuclei to over power energy barrier. |
|
| 19. |
Rutherford studied the first nuclear reaction `[._(7)^(14)N(alpha,p) ._(8)^(17)O]` which take place with a change in energy equivalent to `1.193 MeV`. Later on various types of nuclear reactions such as artifical radioactivity, artifical transmutatoon , nuclear fission, nuclear fussion, spallation reactions etc. were studied. The reaction, `._(33)^(75) As + ._(1)^(2)H rarr ._(25)^(56)Mn + 9 ._(1)^(1)H + 12 ._(0)^(1)n` is called nuclear reaction or:A. Spallation reactionB. Induced radioactivityC. Nuclear fusionD. Artifical radioactivity |
|
Answer» Correct Answer - `a` Spalllation reactions involves emission of large no. of subatomic particles. |
|
| 20. |
Calcualte the energy released (in joule and `MeV`) in the follwing nulcear reaction: `._(1)^(2)H + ._(1)^(2)H rarr ._(2)^(3)He + ._(0)^(1)n` Assume that the masses of `._(1)^(2)H, ._(2)^(3)He` and neutron `(n)` are `2.0141,3.0160` and `1.0087` respectively in amu. |
|
Answer» Correct Answer - `5.223xx10^(3)J, 3.260 MeV` |
|
| 21. |
Statement: Rutherfored studied the firest nuclear reaction: `._(7)^(14)N + ._(2)^(4)He rarr ._(8)^(17)O + ._(1)^(1)H + 1.193 MeV` Explanation: `alpha-` particels lesser than energy `7.6 MeV` were found ineffective.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
|
Answer» Correct Answer - `d` `alpha-` particles with energy lesser than `7.6 MeV` were not capable to penerate nucleaus and over power the repusilve forces. |
|
| 22. |
Calculate the energy released in joules and MeV in the following nuclear reaction : `._(1)^(2) H+_(1)^(2) rarr _(2)^(3)He + _(0)^(1)n` Assume that the masses of `._(1)^(2)H, _(2)^(3)He` and neutron (n) respectively are `2.020, 3.0160 ` and `1.0087` in amu. |
|
Answer» Correct Answer - `[Delta E =2.28 xx 10^(-12)J =14.25 MeV ]` `DeltaE=Delta m xx931.5 Me rArrDeltam=SigmaM_(R)-SigmaM_(P) rArr Deltam=2(2.020)-(3.0160+1.0087)` `rArrDeltaE=Deltamxx931.5=14.25 MeV` |
|
| 23. |
Which nuclear raction is not correct?A. `._(29)^(63)CU(p, ._(1)H^(2)) ._(29)^(62)Cu`B. `._(4)^(9)Be(alpha, ._(0)n^(1)) ._(12)^(6)C`C. `._(5)^(10)B(alpha, ._(0)n^(1)) ._(13)^(7)N`D. `._(27)^(59)Co(._(0)n^(1), ._(1)H^(2)) ._(25)^(62)Mn` |
|
Answer» Correct Answer - `(d)` Equate at no. and mass no. |
|
| 24. |
The radioactivity of a sample is `R_(1)` at a time `T_(1)` and `R_(2)` at a time, `T_(2)` If the half-line of the secimen is `T`, the number of atoms that have disintergrated in the time `(T_(2) - T_(1))` is proportional to:A. `(R_(1) T_(1) - R_(2)T_(2))`B. `(R_(2) - R_(1))`C. `(R_(2) - R_(1))//T`D. `(R_(2) - R_(1))T//0.693` |
|
Answer» Correct Answer - `(d)` `R_(1) = lambda N_(1)`, `R_() = lambda N_(2)`. No, of atoms decayed in time `(T_(2) - T_(1)) = N_(1) - N_(2)` `= (R_(1) - R_(2))/(lambda) = (R_(1) - R_(2))/(0.693) T.` |
|
| 25. |
A radioactive nulei has half-life of `1.0` minute. If one of the nuclie decay now, the next nuclei will decay after:A. `1.0` minuteB. `1//2` minuteC. any timeD. `1//N` minute (`N` is no. of nuclei present at that moment) |
|
Answer» Correct Answer - `(c)` Radioactive decay obey first oder an is instantaneous. |
|
| 26. |
If two light nuclie are fused together in nuclear reaction, the average energy per nucleon:A. increaesB. decreasesC. reamians sameD. none of these |
|
Answer» Correct Answer - `(b)` Mass decay occurs. |
|
| 27. |
Which of the following nuclie have two magic numbers ?A. `._(8)^(16)O`B. `._(2)^(4)He`C. `._(92)^(238)U`D. `._(82)^(208)Pb` |
|
Answer» Correct Answer - `(a,b,d)` `2,8,20,28,50` and `82` are magic no.of proton. `2,8,20,28,50,82` and `126` are magic no.of neutron. |
|
| 28. |
True or False Statements : A fraction `f_(1)` of a radioactive sample decays in one mean life, and a fraction `f_(2)` decays in one half-life, then `f_(1) gt f_(2)`. |
|
Answer» Correct Answer - [T] `(c_(t))/(c_(0))=e^(-lambdax(1)/(lambda))=((1)/(e))=f_(1)=0.36` (left ) then decay `0.633(f_(1))` `(c_(1))/(c_(0))=e^(-lambdax(ln2)/(lambda))=2^(-1)=((1)/(2))=f_(2)=0.5` (left)decay `0.5(f_(2)` `f_(1) gt f_(2)` |
|
| 29. |
A sample contains two radioactive nuclie `x` and `y` with half-lives `2` hour and `1` hour respectively. The nucleus `x`-decays to `y` and `y`-decays into a stable nucelus z.At t = 0, the activates of the components in the same were equal. Find the ratio of the number of the active , nuclei of `y` at `t = 4` hours to the number at `t = 0`. |
|
Answer» Correct Answer - `0.25` |
|
| 30. |
A radioactive substance has 0.1 gm at a particular instant and has an average life of 1 day . The mass of the substance which decays during the 4th day is give by :A. `6.25 mg`B. `12. 5 mg`C. `3.2 mg`D. `1.25 mg` |
|
Answer» Correct Answer - C `[t=(1)/(lambda)=1]`day After 3 days left `w_(1)=wxx e^(-lambdat)` `w_(1)=wxx e^(-3)` After 4 days left `w_(2) =w xx e^(-4)` `w_(1)-w_(2) =we^(-3)=0.1 [ (1)/(e^(3))-(1)/(e^(4))]=3.2 mg ` |
|
| 31. |
A radiioactive element `X`, decays by the sequence and with half-lives, given below: `X["Half-life" = 30 "min"] overset(lambda_(1))(rarr) Y + alpha` `Y["Half-life" = 2 "day"] overset(lambda_(2))(rarr) Z + 2beta` Which of the follwing statements is correct?A. Disinetration constant `lambda_(2) gt lambda_(1)`B. Atomic number of `X` and `Z` are same.C. The mass numner of `X` and `Z` are sameD. `Y` and `Z` are isotopes. |
|
Answer» Correct Answer - `(b)` For `X: t_(1//2) = 30` min, `:. lambda_(1) = (0.693)/(30)` For `Y: t_(1//2) = 2xx24xx60, :. lambda_(2) = (0.693)/(2xx24xx60)` Thus `lambda_(1) gt lambda_(2)` Also emission of one `alpha` and `2beta` give rise to the formation of isotope, i.e, `X` and `Z` have same at no. |
|
| 32. |
Fill in the blanks with appropriate items : A radioactive substance decays 20% in 10 min if at start there are ` 5 xx 10^(20)` atoms present , after ________ hour the number of atoms will be reduced to ` 10^(18)`. |
|
Answer» Correct Answer - `4.65` `k=(1)/(10) ln .(100)/(80) =(1)/(t) "in" (5xx10^(20))/(10^(18)) rArr (1)/(10)"in"(5)/(4)=(1)/(t) "in" 500 rArr (10 xx"in" 500)/("ln"(5)/(4))` `t=278.5` min. `t=4.641` hr. |
|
| 33. |
What is the mean life (in min.) of `Co^(55)` radionuclide, if its activity is known to decrease `4.0%` per hour? The decay product is non-radioactive. [ln 100 =4.6, ln 96=4.56] |
|
Answer» Correct Answer - [1500] `t_("avg")=(1)/(lambda) " " lambda xx 1 =ln(100)/(96) " " lambda=0.04 hr^(-1)` `t_("avg")=(1)/(lambda)=25 xx60 xx=1500` |
|
| 34. |
Fill in the blanks with appropriate items : The half-life period of radioactive element _______ minute if 75% of it disintegrates in 40 min. |
|
Answer» Correct Answer - 20 min. `t_(3//4)=2t_(1//2)=40 =2t_(1//2) rArr t_(1//2)=20 min` |
|
| 35. |
True or False Statements : Half life for certain radioactive element is 15 min. Four nuclei of that element are observed at a certain instant of time. After fifteen minutes, it can be definitely said that two nuclei will be left undecayed. |
|
Answer» Correct Answer - [F] Calculation is applicable on the large amount of ratio active substance and it is propable. |
|
| 36. |
A radioactive sample had an initial activily of 56 dpm (disintegration per min). After `69.3 min` it was found to have an activity of 28 dpm. Find the number of atoms in a sample having an activity of 10 dpm.A. 693B. 1000C. 100D. 10000 |
|
Answer» Correct Answer - B Initial activity = 56 dpm after `69.3` min activity = 28 dpm `:. T_(1//2) =69.3 min " " lambda =(0.693)/(t_(1//2))=(0.693)/(69.3)=10^(-2min)` Activity `=lambda N` `10=10^(-2)xxN " " rArr " " N=1000` |
|
| 37. |
Calculate the specific activity of a radioactive substance `._(98)^(250)Cf`, if its half life is `6.93 min`. Express your answer in terms of `10^(16) dps`. (Use : `N_(A)=6 xx10^(23)`) |
|
Answer» Correct Answer - [0090] Sp. Activity `=lambda N=(1)/(250)xx6xx10^(23)xx(ln2)/(6.93 xx 60)=(1)/(250)xx6xx10^(23)xx(1)/(600)=400xx10^(16)dps` |
|
| 38. |
A radioactive isotope has initial activity of `28` dpm Its activity is reduced to `14` dpm after half an hour. The initial number of nuclide in sample was:A. `200`B. `400`C. `600`D. `1211` |
|
Answer» Correct Answer - `(d)` Rate `= lambda N`, Also `t = (2.303)/(lambda) log ((r_(0))/(r))`, `:. 30 = (2.303)/(lambda) log ((28)/(14))` or `lambda = (2.303)/(30) xx 0.3010`, No `28 = (2.303xx0.3010)/(30) xx N` `:. N = 1211` atoms |
|
| 39. |
The half-life of `.^(32)P` si `14.3` day. Calculate the specific activity of a phosphours containing seciment having `1.0` part per million `.^(32)P` (Atomic weight of `p = 31`) |
|
Answer» Correct Answer - `0.295 Ci` per `g` |
|
| 40. |
Fill in the blanks with appropriate items : The half life of the nuclide `Rn^(220)` is `54.5 sec`., _______ kg of radon is equivalent to 1 millicurie. |
|
Answer» Correct Answer - `1.06 xx 10^(-15)` `A=Nlambda` ` 1xx10^(-3) xx3.7 xx10^(10)` d.p.s `=Nxx(0.693)/(54.5) " " ` 1curie`=3.7 xx 10^(10)` d.p.s. `N=290.98 xx10^(7) rArr (N)/(N_(A))=(29.98 xx10^(7))/(6.023 xx10^(23))=48.31 xx 10^(-16)` moles Mass of Rn`=220 xx48.31 xx 10^(-16)` `rArr 10628.56 xx 10^(-16)` gm `rArr 1.062 xx 10^(-12)gm rArr (1.06 xx 10^(-12))/(1000)kg rArr 1.06 xx 10^(-15) kg` |
|
| 41. |
The human body contains `18%` carbon by weight, in which `.^(14)C` is `1.56xx10^(-6)` per cent. If the half-life of `.^(14)C` is `5570` year, then the number of disintengration per mintute in the body of this weight is:A. `194`B. `1940`C. `19400`D. `28600` |
|
Answer» Correct Answer - `(d)` Amount of `.^(14)C` in body `= (18 xx w xx 1.56xx10^(-10)xx6.023xx10^(23))/(100xx100xx14)` (where `w` is `wt`. Of body) `:.` rate `= lambda xx` No. of atoms `= (0.693)/(5570xx365xx24xx60)` `xx (18xx1.56xx10^(-10)xx6.023xx10^(23))/(100xx100xx14)w` Rate/Body `wt. = 2.86xx10^(4)` dpm |
|
| 42. |
Upon irradiating californium with neutrons, a scientist discovered a new nuclide having mass scientist discovered a new nuclide having mass number of 250 and a half-life of `0.50hr`. Three hours after the irradiation the observated radioactivity due to the nuclide was `10 dis//min`. How many atoms of the nuclide were perepared initially? |
|
Answer» Correct Answer - `2.766xx10^(4)` atoms |
|
| 43. |
Statement: Nuclide `._(13)^(30)Al` is less stable than `._(20)^(40) Ca` Explanation: Nuclides having odd number of protons and neutrons are general unstable.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
|
Answer» Correct Answer - `c` It is a fact and `._(13)^(30) Al` has odd number of `p` and `n` |
|
| 44. |
The radioactive elements `A` and `B` have half-lives of `15` and 5 minute respectively. Thew experiment begins with 4 times the number the number of `B` atoms as `A` atoms. At which of the followig times does the number fo `A` atoms left equals the number of `B` atoms left:A. `30 "minute"`B. `15 "minute"`C. `10 "minute"`D. `5 "minute"` |
|
Answer» Correct Answer - `(b)` For `A: t = (2.303xx15)/(0.693) log ((a)/(x))` For `B: t = (2.303xx5)/(0.693) log ((4a)/(9x))` `= (2.303xx5)/(0.693) [log 4 + log (a)/(x)]` `:. t = 15` minute |
|
| 45. |
Classify each of the following nuclides as either "beta`(._(-1)^(0)beta)` emitter " , or " positron `(._(+1)^(0)beta)` emitter " : `{:(._(20)^(49)Ca,._(5)^(8)B,._(13)^(30)Al,,),(._(80)^(195)Hg,._(67)^(150)Ho,._(36)^(94)Kr,,):}` Note : `._(36)^(84) Kr ,_(80)^(200) Hg ` and `._(67)^(165)Ho` are stable |
|
Answer» Correct Answer - beta emitter : `.^(49)Ca, .^(30)Al, .^(94)Kr,` positron emitter : `.(195)Hg, .^(8)B, .^(150)Ho` `{:(((n)/(p))gt((n)/(p))_("stable") "then"(._(0)^(1)nrarr._(1)^(1)p+._(-1)^(0)e)rarrbeta("emission"),,,,),(((n)/(p)) lt ((n)/(p))_("stable")"then"(._(1)^(1)prarr._(0)^(1)n+._(+1)^(0)e)rarrbeta^(+)("emssion"),,,,),(._(20)^(49)Ca ((n)/(p))gt((n)/(p))_("stable")rarr._(-)beta,,._(80)^(195)Hg((n)/(p))gt((n)/(p))_("stable")rarr._(+)beta,,),(._(5)^(8)B((n)/(p))lt((n)/(p))_("stable")rarr._(+)beta,,._(67)^(150)Hg((n)/(p))lt((n)/(p))_("stable")rarr._(+)beta,,),(._(13)^(30)Al((n)/(p))gt((n)/(p))_("stable")rarr._(-)beta,,._(36)^(94)Kr((n)/(p))lt((n)/(p))_("stable")rarr._(-)beta,,),("beta emitter Ca,Al, Kr",,(._(+)beta)"emitter Hg,B,Ho",,),(,,,,):}` |
|
| 46. |
Nuclides having some atomic number and same mass number but different rate of decay are called:A. isotonesB. isobarsC. nuclear isomersD. isotopes |
|
Answer» Correct Answer - `(c)` It is definiation of nuclear isomers. |
|
| 47. |
Two radioactive nuclides `A` and `B` have half-lives in the ratio `2:3` respectively. An experiment is made with one mole of each of them. Calculate the molar ratio of `A` and `B` after a time interval of three times of half-line of `A`.A. `1:2`B. `2:1`C. `1:3`D. `3:1` |
|
Answer» Correct Answer - `(a)` Half-life of `A` si `2a` and half-line of `B` si `3a` For `A: 6a = (2.303xx2a)/(0.693) log ((1)/(n_(A)))` For `B: 6a = (2.303xx3a)/(0.693) log ((1)/(n_(g)))` `:. N_(A) = 0.125` `N_(B) = 0.250` `:. (N_(A))/(N_(B)) = 1:2` |
|
| 48. |
Two radioactive nuclides `A` and `B` have decay constant `10lambda` and `lambda` respectively. If initially they have same number of nuclei, calculate the ratio of nuclei of `A` and `B` after a time `1//9lambda` |
|
Answer» For `A: N_(1) = N_(o)^(-10lambda t)` For `A: N_(2) = N_(o)e^(-10lambda t)` `:. (N_(1))/(N_(2)) = (e^(-10 lamda t))/(e^(-lambda t)) = e^(-9lambdat) = e^(-9lambda xx1//9lambda) = e^(-1) = (1)/(e)` |
|
| 49. |
Two radioactive nuclides A and B have half-lives 50 min and 10 min respectively . A fresh sample contains the nuclides of B to be eight times that of A. How much time should elapse so that the mumber of nuclides of A becomes double of B ?A. 30B. 40C. 50D. None |
|
Answer» Correct Answer - C Initial no. of nuclei`{:(Aoverset(t_(1//2)=50 min)rarrX,,B overset(t_(1//2)=10 min)rarrX),(N_(0),,8N_(0)),(,,),(,,):}` Let after n `t_(1//2)` of A, no. of nuclei of A becomes double of B. `N_(A)=(N_(0))/(2^(n)) " " N_(B)=(8N_(0))/(2 ^(5n)) " " ( . :. "in" n t _(1//2) of A, t_(1//2) of B " will " 5n)` `N_(A)=2 N_(B)` `(N_(0))/(2^(n))=2 xx (8N_(0))/(2^(5n))` `2^(4n)=16 = 2^(4) rArr n=1` After one half life time of A (i.e. 50 min) No. of nuclei of A will become double of B. |
|
| 50. |
A piece of wood from an archelogical source shows a `.^(14)C` activity which is `60%` of the activity found in fresh wood today. Caculate the age of the archeologival sample. (`t_(1//2)` for `.^(14)C = 5570` year) |
|
Answer» `t = (2.303)/(lambda) log (N_(0))/(N)` (`r_(0) prop N_(0)` and `r prop N`) `= (2.303xx5770)/(0.693)` log `(100)/(60)` `= 4253 yr, [:. (r_(0))/(r) = (N_(0))/(N) = (100)/(60)]` |
|