94420 + Interview Questions in Chemistry in Class 12 Page 1758 InterviewSolution

87851.	(A) Rate of reaction can also increase with the formation of product, if one of the products acts as a catalyst. (R ) A catalyst lowers the activation energy of the reaction
Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false Answer :B

Discussion

87852.	Assertion :Rate of evaporation increases with an increase in the surface area of the vessel Reason :Evaporation is a surface phenomenon
Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE (A) is false but (R) is true Answer :A

Discussion

87853.	A: Rate constant of a reaction at a particular temperature is constant R: The value of rate constant K is indepdnent of initial concentration.
Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false Answer :B

Discussion

87854.	(A) Rate of constant of a reaction increases with increase in temperature (R ) Increase in temperature increases in the number of activated molecules
Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false Answer :A

Discussion

87855.	Rate constant of reaction at 300 K and 400 K are 0.0345 S^(-1) and 0.1365 S^(-1) respectively. Calculate the activation energy for the reaction. [Given : R = 8.314 J K^(-1) mol^(-1)]
Answer» Solution :(A) `logK_(2)/K_(1) = E_(a)/2.303R[T_(2)-T_(1)/T_(1)T_(2)]` `log0.136/0.034 = E_(a)/2.303 xx 8.314[400-300/300xx400]` `E_(a)= 13.8 kJ//mol` (b) Consider a zero order reaction `R rarr P` Zero order reaction is one in which rate is proportional to zeroth power of reactant concentration. According to rate law for zero order reaction, `Rate = k[R]^(0)`Rate = `k xx 1` where k is rate CONSTANT or velocity constant. But rate is defined by, Rate `= -d[R]/dt:.-d[R]/dt = k` Rearranging the equation, we get`d[R] = -kdt` On integration ` int d[R] = -kint dt[R] = -kt + I` where I is called integration constant To get, I, when t = 0,`[R] = [R]_(0)` the initial concentration of the reactant `[R]_(0) = -k xx 0 + II = [R]_(0)` Substitute I VALUE in equation(1), we get `[R] = -kt + [R]_(0)[R]_(0)-[R] = kt` `k = [R]_(0) - [R]/t` On rearranging, `[R] = -kt + [R]_(0)` This equation is in y = mx + c FORM and suggests that A plot of [R] versus .t. gives a STRAIGHT LINE with a slope = -k and y-intercept =` [R]_(0)`

Discussion

87856.	(a) Rate constant of a first order reaction is 0.0693 mi n^(-1). Calculate the percentage of the reactant remaining at the end of 60 minutes.
Answer» Solution :`K=(2.303)/(t)"LOG"(R_(0))/(R)` `0.0693=(2.303)/(16)"log"(100)/(R)` `R=1.56%`

Discussion

87857.	(a) Rate constant of a chemical reaction rises to double by increase in temperature of 10^(@)C. Explain with labelled distribution curve. (b) The rate constant of a first order reaction increases by four times when the temperature changes from 350 K to 400 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature (R=8.314" JK"^(-1)"mol"^(-1),log4=0.6021)
Answer» SOLUTION :`32.27" KJ MOL"^(-1)`

Discussion

87858.	A rarr B + C Time ""t " "prop Total pressure of (B + C) " "P_(2) " "P_(3) Find k.
Answer» <P> SOLUTION :`K=(1)/(t)" In "(P_(3))/((P_(3)-P_(2)))`

Discussion

87859.	A rarr (CH_(3))_(2)C= CHCOCH_(3), A is
Answer» Acetone Acetaldehyde Propionaldehyde Formaldehyde Solution :Two MOLECULES of acetone CONDENSE in presence of dry `HCI` GOES to FORM mesityl oxide.

Discussion

87860.	Ararr B + C Time 0 t Volume of reagent V_(1)" " V_(2) The reagent reacts with A, B and C. Find k.
Answer» Solution :`K=(1)/(t)" In "(V_(1))/((2V_(1)-V_(2)))`

Discussion

87861.	Arrange the following solutions in increasing order of their osmotic pressures . (i)34.2 g/litre surcrose (ii)60 g/litre of urea (iii)90 g/litre of glucose (iv)58.5 g/litre of sodium chloride
Answer» Solution :`"Osmotic pressure of a solution Is calculated with the help of Van't Hoff equation":pi=I MRT` (a) `M=((34.2"g L"^(-1)))/((342"g mol"^(-1)))=0.1"mol L"^(-1)=0.1M` `pi=i MRT=(1)XX(0.1M)xxRT=(0.1M)xxRT` (b) `M=((60.0"g L"^(-1)))/((60"g mol"^(-1)))=0.1" mol L"^(-1)=1M` `pi=i MRT=(1) xx(1 M)xxRT=(1M)xxRT` (c) `M=((90.0"g L"^(-1)))/((180"g mol"^(-1)))=0.5" g mol"^(-1)=0.5 M` `pi = i MRT = (1)xx(0.5M)xxRT=(0.5M)xxRT` (d) `M=((58.5"g L"^(-1)))/((58.5"g mol"^(-1)))=1" mol L"^(-1)=1M` `pi=iMRT=(2)xx(1M)xxRT=(2M)xxRT.` `"CORRECT increasing order of osmotic pressure "= (a)lt (c) lt (b) lt (d).`

Discussion

87862.	A rarr 2B + 3C Time ""t " "prop Volume of reagent v_(2) " "v_(3) Reagent reacts with all A, B and C. Find K.
Answer» SOLUTION :`K=(1)/(t)" In "(4V_(3))/((5V_(3)-V_(2)))`

Discussion

87863.	A raduction of atomic size with increase in atomic number is a characteristic of elements of
Answer» radioactive series HIGH atomic MASS f-block all of these Answer :C

Discussion

87864.	Aradium Ra_88^224isotope, on emission of an alpha-particle gives rise to a new element whose mass number and atomic number will be:
Answer» 220 and 86 225 and 87 228 and 88 224 and 86 Solution :When an `alpha`-particle is emitted by any nucleus then atomic WEIGHT decrease by four UNITS and atomic number DECREASES by two units `._(88)RA^(224) overset(-alpha)rarr ._(86)X^(220)`

Discussion

87865.	(A): Radium is a natural source of two noble gases (R): Radium undergoes alpha-emission spontaneously
Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is false Both (A) and (R) are false Answer :A

Discussion

87866.	A radioisotope undergoes decomposition which follows two parallel paths as shown below The percentage distribution of 'B' and 'C' are
Answer» 80% of 'B'and 20% of 'C' `76.83%'B' and 23.17%'C'` 90%'B' and 10% 'C' 60% 'B' and 40% 'C' Solution :The percent distribution of 'B' is `(k_(1))/(k_(1)+K_(2))XX100 =76.83%, "" 'c'" is" (k_(2))/(k_(1)+k_(2))XX 100=23.17%`

Discussion

87867.	A radioelement has atomic number 90 and mass number 232. What is the atomic number and mass number of the end product obtained by loss of 6 - alpha and 4 - betaparticles?
Answer» 82208 88206 82206 84, 204 Solution :The nuclear reaction is `""_(90)X^(232) to ""_(Z)Y^(A) + 6 ""_(2)alpha^(4) + 4 ""_(1)beta^(0)` Considering MASS number, `232 = A + 24 implies A = 208` Considering atomic number, `90 = Z + 12 - 4 implies Z = 82` Now atomic number and mass number of the end PRODUCT is 82 and 208 respectively.

Discussion

87868.	A radioisotope has a half-life of 900 seconds. Calculate the fraction of the original isotope which will remain behind after four half-life periods.
Answer» SOLUTION :`1//16`

Discussion

87869.	A radioisotope has a …….. Of 10 days. If today 125 g of it is left, what was its weigh 40 dayscarlicr?
Answer» 600 g 1000 g 1250 g 2000G SOLUTION :`n = 40/10 = 4, 125 g = 1/(2^4) [A_0]` or `[A_0] = 2000g. `

Discussion

87870.	A radioactive substance having a half life period of 5 days was received after 20 days. It was found that there was 3 g of the isotope in the container. The initial weight of the isotope when placed was:
Answer» 12 g 24 g 36 g 48 g Answer :D

Discussion

87871.	A radioactive substance takes 20 min to decay 25%. How much time will be taken to decay 75%
Answer» 96.4 min 68 min 964 min 680 min Solution :For 25% DECAY `K = (2.303)/(20) "log"(100)/(75) = (2.303)/(20) xx 0.1249 = 0.1438` For 75% decay `t = (2.303)/(0.01438) "log" (100)/(25) = 96.4` MINUTE

Discussion

87872.	A radioactive substance is decaying with t_((1)/(2))=30 days. On being separated into two fractions, one of the fractions, immediately after separation, decays with t_(1//2) = 2 days. The other fraction, immediately after separation, would show
Answer» CONSTANT activity increasing activity decay with `t_(1//2)=30` days decay with `t_(1//2)=28` days Answer :B

Discussion

87873.	A radioactive substance (parent ) decays to its daughter element the age of radioactive substance (t) is related to the daughter (d) parent (p) ratio by the equation :
Answer» <P>`t = (1)/(lambda) "ln" (1 + (p)/(d))` `t = (1)/(lambda) "ln" (1 + (d)/(p))` `t = (1)/(lambda) "ln" ((d)/(p))` `t = (1)/(lambda) "ln" ((p)/(d))` Solution :`A to D ` `(N_(0) - X) = N^(PX) = N^(D) , (N_(0) - x) = N^(p) , N_(0) - N^(D) = N^(p) , N_(0) = (N^(D) + N^(p))` `t = (1)/(lambda) XX log ((N^(D) +N^(P))/(N^(P))) , t = (1)/(lambda) xx 1_(n) (1 + (N^D)/(N^P))`

Discussion

87874.	A radioactive substance is reduced to 1/8 of its original concentration in 24 s. The rate constant of the reaction is :
Answer» `(0.69)/(16) s^(-1)` `(1)/(8)s^(-1)` `(1)/(24)s^(-1)` `("In"2)/(8)s^(-1)` Solution :(D ) `([A]_(0))/(2^(N)) = ([A]_(0))/(8) " ""or"" " 2^(n) =8 , n =3 ` `t_(1//2) = (24)/(3) = 8s` `K = (In2)/(8) s^(-1)` .

Discussion

87875.	A radioactive substance has a halflife of 50 days . Fraction of the material left behind after 100 day will be :
Answer» 0.5 0.75 0.125 0.25 Solution :`I overset(50 "DAYS")(to)(1)/(2)overset(50 "days") (to) (1)/(4)`

Discussion

87876.	A radioactive substance has t_(1//2) 60 minutes. After 3 hrs, what precentage of radioactive substance will remain
Answer» 0.5 0.75 0.25 0.125 Solution :AMOUNT left `= (N_(0))/(2^(3)) = (100)/(8) = 12.5%`

Discussion

87877.	A radioactive substance has t_(1//2) 60 minutes. After 3 hrs, what percentage of radioactive substance will remain ?
Answer» 0.5 1.75 0.25 0.125 Solution :No. of HALF lives in 3 HRS =3 `:.C_(N)=(1)/(2^(n))=(1)/(2^(3))=(1)/(8)` `:.` AMOUNT LEFT =12.5%

Discussion

87878.	A radioactive substance having a half life period of 3 days was received after 12 days. It was found that there was 3 g of the isotope in the container. The initial weight of the isotope when placed was :
Answer» 12 g 24 g 36 g 48 g Answer :D

Discussion

87879.	Carbon-14 has a half life period of 5760 years. 100 mg of sample containing C-14 is reduced to 25 mg in:
Answer» 11520 YEARS 2880 years 1440 years 17280 years. ANSWER :A

Discussion

87880.	A radioactive substance has a constant activity of 2000 disintegrations/minute. The material is separated into two fractions one of which has an initial activity of 1000 disintegrations per minute while the other fraction decays with t_((1)/(2))=24 hours. The total activity in both samples after 48 hours of separation is
Answer» 1500 1000 1250 2000 Answer :D

Discussion

87881.	A radioactive substance has 0.1gm at a particuular instant aand has an average life of 1 day. The mass of the substance which always during the 4th day is given by:
Answer» 6.25mg 12.5mg 3.15mg 1.25mg Answer :C

Discussion

87882.	A radioactive sample of C^(14) showing 20 curie activity. Calculate its activity ("in" C_(i)) after 12000 yrs (t_(1//2)"of" C^(14) = 6000 "yrs")
Answer» ANSWER :5

Discussion

87883.	A radioactive substance .)(88)X^(228) (IIA) emits 3alpha and 3beta-particles to form 'Y'. O which group of long form of the periodic table does 'Y' belong?
Answer» IVA VA VIA VIIA Solution :`UNDERSET("IIA GROUP")(._(88)X^(228)) to underset("VIIA group")(._(85)Y^(216))+3alpha+3beta`

Discussion

87884.	A radioactive sample had an initial activity of 56 dpm . After 69.3 minutes , it was found to have an activity of 28 dpm , the number of atoms in a sample having an activity of 100 dpm is 10^(x) . The value of x is …….
Answer» Solution :56 dpm `OVERSET(t_(1//2) = 69.3 "MIN") (to) 28 dpm A = lambda N , 100 = ((0.693)/(69.3)) , N_(t) , N_(t) = 10^(4)`

Discussion

87885.	A radioactive sample has a half-life of 1500 years. A sealed tube containing 1 gm of the sample will contain after 3000 years
Answer» 1 G of the sample 0.5 g of the sample 0.25 g of the sample 0.00 g of the sample Solution :`T = t_(1//2) xx n, :. 3000 = 1500 XXN :. N =2` `:.` Amount left `= (1)/(2^(2)) = (1)/(4) = 0.25 g`

Discussion

87886.	A radioactive sample emit n beta - particles is 2 sec , In next 2 sec it emits 0.75 n beta- particle , what is the mean life of the sample?
Answer» Solution :LET initial number of nuclei `=N_(0)` `N_(0)overset(2 "sec")rarr (N_(0)-N) overset(2 "sec")rarr(N_(0)-n-0.75 n)` (No. of `beta` particles emitted=No. of nuclei disintegrated) No. of nuclei disintegrated in time is given by `=N_(0)(1-e^(-lambdat t)` `:.` for initial 2 seconds `n=N_(0)(1-e^(-2lambda)) "" `...........(i) for next2 seconds `0.75 n=(N_(0)-n)(1-e^(-2 lambda)) "" ` ..........(ii) Subtracting (ii) from (i) ((i)-(ii)) `.25 n=n (1-e^(-2lambda))` `e^(-2lambda)=0.75` `-2lambda=ln. (3)/(4) "" rArr 2lambda =ln .(4)/(3)` `t_("avg")=(1)/(lambda)=(2)/(ln((4)/(3)))=6.95 "sec"`

Discussion

87887.	A radioactive sample has half life of 1500 years. A sealed tube containing 1 g of a sample will contain ... g of the sample after 3000 years. The missing figure is :
Answer» 1 G of the SAMPLE 0.5 g of the sample 0.25 g of the sample 0.00 g of the sample. Answer :C

Discussion

87888.	A radioactive nuclide is produced at a constant rate of alpha- per second. It decay constant is lambda . If N_(0) be the no. of nuclei at time t = 0 , then maximum no. of nuclei possible are
Answer» `( ALPHA)/( LAMBDA)` `N_(0) + ( alpha)/( lambda)` `N_(0)` `( lambda)/( alpha) + N_(0)` ANSWER :A

Discussion

87889.	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)) is
Answer» 0.027 `0.270 xx 10^(-5)` 0.0027 0.00027 Solution :`(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)`

Discussion

87890.	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 is
Answer» `0.10 "MIN"^(-1)` `0.01 "min"^(-1)` `1.0 "min"^(-1)` `0.001 "min"^(-1)` SOLUTION :`K = (0.693)/(t_(1//2)) = (0.693)/(6.93) = 0.10 "min"^(-1)`

Discussion

87891.	A radioactive nuclide generally disintegrates by alpha-emission when its N/P ratio is
Answer» LESS than 1 equal to one equal to 1.2 greater than 1.5 Answer :D

Discussion

87892.	A radioactive nuclide is produced at a constant rate alpha-"per second"/ It's decay constant is lambda sec^(-1). If N_(0) b the number of nuclei at time t=0, then maximum number of nuclei at any instant possible are:
Answer» `(ALPHA)/(LAMBDA)` `N_(0)-(alpha)/(lambda)` `(lambda)/(alpha)` `alpha+lambdaN_(0)` ANSWER :A

Discussion

87893.	A radioactive mixture contaning a short lived species A and short lived species B. both emitting alpha-"particles" at a given instant, emit at rate 10,000 alpha-"participle" per minute. 10minutes later, it emits at the rate of 7000 particles per minutes. It half lives of the species are 10minutes and 100 hours respectively, then the ratio of activities of A : Bin the initial mixture was:
Answer» `3: 7 ` `4:6` `6:4` `10:4` ANSWER :C

Discussion

87894.	A radioactive nucleus decays by emitting one alpha and two beta particles, the daughter nucleus is ………………….. of the parent
Answer» SOLUTION :ISOTOPE

Discussion

87895.	A radioactive nuclide emitts gamma- rays due to the :
Answer» emission of an ELECTRON from its orbital nuclear ENERGY TRANSITION from a higher state to a lower state presence of LESS NEUTRONS than protons presence of more neutrons than protons Answer :B

Discussion

87896.	A radioactive isotopeA undergoes simultaneous decay to two different nuclei as Assuming that initially neither P norQ was present, after how many hours, amount of Q will be just double to the amount of A remaining?
Answer» Both Assertion and REASON are CORRECT and Reason is the correct explanation of the Assertion Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion Assertion is correct but Reason is INCORRECT Assertion is incorrect but Reason is correct Answer :6

Discussion

87897.	A radioactive material exhibits an intensity of 3.8 mCi at time 't' & an intensity of 1.35 m Ci five minutes later. Calculate the halflife period of the substance .
Answer» 202.28 SEC 102.28 Sec 3.4 minutes 34 minutes SOLUTION :`((0.693)/(t_(1//2))) xx 5 = 2.303 LOG ((3.8)/(1.36)) , t_(1//2) = (0.693 xx 5)/(2.303 xx 0.446) = 3.374` min = 3.4 min

Discussion

87898.	A radioactive isotope having a half-life of 3 days was received after 12 days. It was found that there were 3g of the isotope in the container. The initial weight of the isotope when packed was
Answer» 12g 24g 36g 48g Answer :D

Discussion

87899.	A radioactive isotope having a half-life of 3 days was received after 12 days. It was found that there were 3 g of the isotope in the container. The initial weight of the isotope when packed was
Answer» 12 G 24 g 36 g 48 g Solution :`n = (12)/(3) = 4` `:. N_(0) = N xx 2^(n) = 3 xx 2^(4) = 48 g`

Discussion

87900.	A radioactive isotope having a half life of 2.3 days was received after 9.2 days. It was found that there were 300 mg of the isotope in the container. The initial amount of the isotope was
Answer» 3600 MG 1200 mg 2400 mg 4800 mg. Solution :9.2 days = 4 half-lives Amount LEFT after N half-lives = `1/(2^n) XX a` `300 = 1/(2^4) a` or `a = 300 xx 16 = 4800` mg.

Discussion

Explore topic-wise InterviewSolutions in .

(A) Rate of reaction can also increase with the formation of product, if one of the products acts as a catalyst. (R ) A catalyst lowers the activation energy of the reaction

Assertion :Rate of evaporation increases with an increase in the surface area of the vessel Reason :Evaporation is a surface phenomenon

A: Rate constant of a reaction at a particular temperature is constant R: The value of rate constant K is indepdnent of initial concentration.

(A) Rate of constant of a reaction increases with increase in temperature (R ) Increase in temperature increases in the number of activated molecules

Rate constant of reaction at 300 K and 400 K are 0.0345 S^(-1) and 0.1365 S^(-1) respectively. Calculate the activation energy for the reaction. [Given : R = 8.314 J K^(-1) mol^(-1)]

(a) Rate constant of a first order reaction is 0.0693 mi n^(-1). Calculate the percentage of the reactant remaining at the end of 60 minutes.

A rarr B + C Time ""t " "prop Total pressure of (B + C) " "P_(2) " "P_(3) Find k.

A rarr (CH_(3))_(2)C= CHCOCH_(3), A is

Ararr B + C Time 0 t Volume of reagent V_(1)" " V_(2) The reagent reacts with A, B and C. Find k.

Arrange the following solutions in increasing order of their osmotic pressures . (i)34.2 g/litre surcrose (ii)60 g/litre of urea (iii)90 g/litre of glucose (iv)58.5 g/litre of sodium chloride

A rarr 2B + 3C Time ""t " "prop Volume of reagent v_(2) " "v_(3) Reagent reacts with all A, B and C. Find K.

A raduction of atomic size with increase in atomic number is a characteristic of elements of

Aradium Ra_88^224isotope, on emission of an alpha-particle gives rise to a new element whose mass number and atomic number will be:

(A): Radium is a natural source of two noble gases (R): Radium undergoes alpha-emission spontaneously

A radioisotope undergoes decomposition which follows two parallel paths as shown below The percentage distribution of 'B' and 'C' are

A radioelement has atomic number 90 and mass number 232. What is the atomic number and mass number of the end product obtained by loss of 6 - alpha and 4 - betaparticles?

A radioisotope has a half-life of 900 seconds. Calculate the fraction of the original isotope which will remain behind after four half-life periods.

A radioisotope has a …….. Of 10 days. If today 125 g of it is left, what was its weigh 40 dayscarlicr?

A radioactive substance having a half life period of 5 days was received after 20 days. It was found that there was 3 g of the isotope in the container. The initial weight of the isotope when placed was:

A radioactive substance takes 20 min to decay 25%. How much time will be taken to decay 75%

A radioactive substance is decaying with t_((1)/(2))=30 days. On being separated into two fractions, one of the fractions, immediately after separation, decays with t_(1//2) = 2 days. The other fraction, immediately after separation, would show

A radioactive substance (parent ) decays to its daughter element the age of radioactive substance (t) is related to the daughter (d) parent (p) ratio by the equation :

A radioactive substance is reduced to 1/8 of its original concentration in 24 s. The rate constant of the reaction is :

A radioactive substance has a halflife of 50 days . Fraction of the material left behind after 100 day will be :

A radioactive substance has t_(1//2) 60 minutes. After 3 hrs, what precentage of radioactive substance will remain

A radioactive substance has t_(1//2) 60 minutes. After 3 hrs, what percentage of radioactive substance will remain ?

A radioactive substance having a half life period of 3 days was received after 12 days. It was found that there was 3 g of the isotope in the container. The initial weight of the isotope when placed was :

Carbon-14 has a half life period of 5760 years. 100 mg of sample containing C-14 is reduced to 25 mg in:

A radioactive substance has 0.1gm at a particuular instant aand has an average life of 1 day. The mass of the substance which always during the 4th day is given by:

A radioactive sample of C^(14) showing 20 curie activity. Calculate its activity ("in" C_(i)) after 12000 yrs (t_(1//2)"of" C^(14) = 6000 "yrs")

A radioactive substance .)(88)X^(228) (IIA) emits 3alpha and 3beta-particles to form 'Y'. O which group of long form of the periodic table does 'Y' belong?

A radioactive sample had an initial activity of 56 dpm . After 69.3 minutes , it was found to have an activity of 28 dpm , the number of atoms in a sample having an activity of 100 dpm is 10^(x) . The value of x is …….

A radioactive sample has a half-life of 1500 years. A sealed tube containing 1 gm of the sample will contain after 3000 years

A radioactive sample emit n beta - particles is 2 sec , In next 2 sec it emits 0.75 n beta- particle , what is the mean life of the sample?

A radioactive sample has half life of 1500 years. A sealed tube containing 1 g of a sample will contain ... g of the sample after 3000 years. The missing figure is :

A radioactive nuclide is produced at a constant rate of alpha- per second. It decay constant is lambda . If N_(0) be the no. of nuclei at time t = 0 , then maximum no. of nuclei possible are

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)) is

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 is

A radioactive nuclide generally disintegrates by alpha-emission when its N/P ratio is

A radioactive nuclide is produced at a constant rate alpha-"per second"/ It's decay constant is lambda sec^(-1). If N_(0) b the number of nuclei at time t=0, then maximum number of nuclei at any instant possible are:

A radioactive nucleus decays by emitting one alpha and two beta particles, the daughter nucleus is ………………….. of the parent

A radioactive nuclide emitts gamma- rays due to the :

A radioactive isotopeA undergoes simultaneous decay to two different nuclei as Assuming that initially neither P norQ was present, after how many hours, amount of Q will be just double to the amount of A remaining?

A radioactive material exhibits an intensity of 3.8 mCi at time 't' & an intensity of 1.35 m Ci five minutes later. Calculate the halflife period of the substance .

A radioactive isotope having a half-life of 3 days was received after 12 days. It was found that there were 3g of the isotope in the container. The initial weight of the isotope when packed was

A radioactive isotope having a half-life of 3 days was received after 12 days. It was found that there were 3 g of the isotope in the container. The initial weight of the isotope when packed was

A radioactive isotope having a half life of 2.3 days was received after 9.2 days. It was found that there were 300 mg of the isotope in the container. The initial amount of the isotope was