Q: For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is :

Number of nuclei remaining =600−450=150 N0N=(21)n600150=(21)t (21)2=(21)t1/2t t=2t1/2=2×10 =20 minute

Q: the half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is

Ratio of remaining nuclei (N) to initial nuclei (No) having half life period of T=30 min at time t is given by :-N0N=(21)t/T⇒t=−Tlog2(N0N)(i) when 40% decay completes in time t1 (i.e N0N0.6 ) [40% decay means, 60% remaining ]⇒t1=−Tlog2(0.6)(ii) when 85% decay completes in time t2(N0N=0.15)t2=−Tlog2(0.15)So time taken,Δt=t2−t1=T[log2(0.6)−log2(0.15)]=Tlog2(0.150.6)=30×log2(4)Δt=60 min

Q: Q.218. The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half-life.

If the half-life is T days (2)−7/T=2⋅51 Hence T7=ln2ln2⋅5 or T=ln2⋅57ln2=5.30 days.

Q: Half-life of a radioactive substance is T. At time t1 activity of a radioactive substance is R1 and at time t2 it is R2. Find the number of nuclei decayed in this interval of time.

Half-life is given by t1/2=λln2 λ=t1/2ln2=Tln2 Activity R=λN ∴N=λR=ln2RT When activity is R1, numbers of nuclei are N1=ln2R1T Similarly, N2=ln2R2T ∴ Numbers decayed =N1−N2=ln2(R1−R2)T

Q: Two radioactive materials X1 and X2 have decay constants 10λ and λ, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time

Nx2(t)Nx1(t)=e1N0e−λtN0e−10λt=e1 (Initially, both have same number of nuclei say N0 )

Q: A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is

From R=R0(21)nwe have, 1=64(21)nor n=6= number of half-lives∴t=n×tt1/2=6×2=12 h

Q: A radioactive sample S1 having an activity of 5μCi has twice the number of nuclei as another sample S2 which has an activity of 10μCi. The half-lives of S1 and S2 can be

Activity of S1=21or λ1N1=21(λ2N2)or λ2λ1=2N1N2or T2T1=N22N1(T= half − life =λln2)Given, N1=2N2∴T2T1=4∴ Correct option is (a).

Q: Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is

After two half-lives 41 th fraction of nuclei will remain un-decayed. Or, 43 th fraction will decay.Hence, the probability that a nucleus decays in two half-lives is 43.

Q: The half-life of 215 At is 100μs. The time taken for the activity of a sample of 215 At to decay to 161 th of its initial value is

R=R0(21)nHere, R= activity of radioactive substance after n half-lives =16R0 (given)Substituting in Eq. (i), we get n=4∴t=(n)t1/2=(4)(100μs)=400μs

Question 1

For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is :

Accepted Answer

Number of nuclei remaining =600−450=150 ​N0​N​=(21​)n600150​=(21​)t​ (21​)2=(21​)t1/2​t​ t=2t1/2​=2×10 =20  minute

Question 2

the half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is

Accepted Answer

Ratio of remaining nuclei (N) to initial nuclei (No​) having half life period of T=30 min at time t is given by :-N0​N​=(21​)t/T⇒t=−Tlog2​(N0​N​)(i) when 40% decay completes in time t1​ (i.e N0​N​0.6 ) [40% decay means, 60% remaining ]⇒t1​=−Tlog2​(0.6)(ii) when 85% decay completes in time t2​(N0​N​=0.15)t2​=−Tlog2​(0.15)So time taken,Δt=t2​−t1​=T[log2​(0.6)−log2​(0.15)]=Tlog2​(0.150.6​)=30×log2​(4)Δt=60 min

Question 3

Q.218. The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half-life.

Accepted Answer

If the half-life is T days
(2)−7/T=2⋅51​
Hence T7​=ln2ln2⋅5​
or T=ln2⋅57ln2​=5.30 days.

Question 4

Half-life of a radioactive substance is T. At time t1​ activity of a radioactive substance is R1​ and at time t2​ it is R2​. Find the number of nuclei decayed in this interval of time.

Accepted Answer

Half-life is given by
t1/2​=λln2​
λ=t1/2​ln2​=Tln2​
Activity R=λN
∴N=λR​=ln2RT​
When activity is R1​, numbers of nuclei are N1​=ln2R1​T​
Similarly,
N2​=ln2R2​T​
∴ Numbers decayed
=N1​−N2​=ln2(R1​−R2​)T​

Question 5

Two radioactive materials X1​ and X2​ have decay constants 10λ and λ, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1​ to that of X2​ will be 1/e after a time

Accepted Answer

Nx2​​(t)Nx1​​(t)​=e1​N0​e−λtN0​e−10λt​=e1​ (Initially, both have same number of nuclei say N0​ )

Question 6

A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is

Accepted Answer

From R=R0​(21​)nwe have, 1=64(21​)nor n=6= number of half-lives∴t=n×tt1/2​​=6×2=12 h

Question 7

A radioactive sample S1​ having an activity of 5μCi has twice the number of nuclei as another sample S2​ which has an activity of 10μCi. The half-lives of S1​ and S2​ can be

Accepted Answer

Activity of S1​=21​or λ1​N1​=21​(λ2​N2​)or λ2​λ1​​=2N1​N2​​or T2​T1​​=N2​2N1​​(T= half − life =λln2​)Given, N1​=2N2​∴T2​T1​​=4∴ Correct option is (a).

Question 8

Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is

Accepted Answer

After two half-lives 41​ th fraction of nuclei will remain un-decayed. Or, 43​ th fraction will decay.Hence, the probability that a nucleus decays in two half-lives is 43​.

Question 9

After 280 days, the activity of a radioactive sample is 6000dps. The activity reduces to 3000dps after another 140 days. The initial activity of the sample in dps is

Accepted Answer

Activity reduces from 6000dps to 3000dps in 140 days. It implies that half-life of the radioactive sample is 140 days. In 280 days (or two half-lives) activity will remain 41​ th of the initial activity. Hence, the initial activity of the sample is 4×6000dps=24000dpsTherefore, the correct option is (d).

Question 10

The half-life of 215 At is 100μs. The time taken for the activity of a sample of 215 At to decay to 161​ th of its initial value is

Accepted Answer

R=R0​(21​)nHere, R= activity of radioactive substance after n half-lives =16R0​​ (given)Substituting in Eq. (i), we get n=4∴t=(n)t1/2​=(4)(100μs)=400μs

Question 11

The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20 th of the radon sample will remain undecayed, is (given log10​e=0.4343 )

Accepted Answer

Using N=N0​e−λtwhere, λ=t1/2​ln2​=3.8ln2​∴20N0​​=N0​e−3.8ln2​tSolving this equation with the help of given data we findt=16.5 days ∴ Correct option is (b).

Question 12

A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity (t=0) of 40μCi. Calculate the number of nuclei that decay in the time interval between t1​=10.0 h and t2​=12.0 h

Accepted Answer

R0​=λN0​⇒N0​=λR0​​
where, λ=t1/2​ln2​
N=N0​e−λt
Find N1​=N0​e−λt1​
and N2​=N0​e−λt2​
∴ Number of nuclei decayed in given
 time =N1​−N2​

Question 13

A freshly prepared sample of a certain radioactive isotope has an activity of 10mCi. After 4.0 h its activity is 8.00mCi.
(a) Find the decay constant and half-life
(b) How many atoms of the isotope were contained in the freshly prepared sample?
(c) What is the sample's activity 30.0 h after it is prepared?

Accepted Answer

(a) R=R0​e−λt
R0​=20MCI
R=8MCI
t=4.0 h
Find λ.
(b) R0​=λN0​. Find N0​=λR0​​
(c) Find R=R0​e−λt

Question 14

Two radioactive elements X and Y have half-life periods of 50 minutes and 100 minutes, respectively. Initially, both of them contain equal number of atoms. Find the ratio of atoms left NX​/NY​ after 200 minutes.

Accepted Answer

In 200 minute time,
n1​= number of half-lives of X
=50200​=4
n2​= number of half-lives of Y
=100200​=2
NY​NX​​=N0​(1/2)2N0​(1/2)4​=41​

Question 15

At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 s the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei,
(b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number.

Accepted Answer

(a) In 10 s, number of nuclei has been reduced to half (25% to 12.5%). Therefore, its half-life is
t1/2​=10 s
Relation between half-life and mean life is
tmean ​=ln2t1/2​​=0.69310​ s
tmean ​=14.43 s
(b) From initial 100\% to reduction till 6.25%, it takes four half-lives.
100%⟶t1/2​​50%⟶t1/2​​25%⟶t1/2​​12.5%⟶t1/2​​6.25%
t=4t1/2​=4(10)s=40 s
t=40 s

Question 16

Natural uranium is a mixture of three isotopes 92234​U,92235​U and 92238​U with mass percentage 0.01%,0.71% and 99.28% respectively. The half-life of three isotopes are 2.5×105yr,7.1×108yr and 4.5×109yr respectively.Determine the share of radioactivity of each isotope into the total activity of the natural uranium.

Accepted Answer

Let R1​,R2​ and R3​ be the activities of U234,U235 and U238 respectively. Total activity,Share of U234,R=R1​+R2​+R3​RR1​​=λ1​N1​+λ2​N2​+λ3​N3​λ1​N1​​Let m be the total mass of natural uranium.Then,Now,m1​=1000.01​m,m2​=1000.71​m and m3​=10099.28​mN1​=M1​m1​​,N2​=M2​m2​​ and N3​=M3​m3​​ where M1​,M2​ and M3​ are atomic weights.∴ RR1​​=M1​m1​​T1​1​+M2​m2​​⋅T2​1​+M3​m3​​⋅T3​1​(M1​m1​​)T1​1​​

Question 17

87​ th of the active nuclei present in a radioactive sample has decayed in 8 s. The half-life of the sample is

Accepted Answer

Number of nuclei left =81​ thNow, 81​=(21​)n∴n= number of half-lives =33 half-lives =8 s∴1 half-life =38​ s

Question 18

Determine the age of ancient wooden items if it is known that the specific activity of 14C nuclide in them amounts to 3/5 of that in lately felled trees. The half life of 14C nuclei is 5570 years. [ Take ln0.6=−0.51]

Accepted Answer

In old wooden atoms the number C14 nuclei steadily decreases because of radioactive decay. (In live tree biological processes keep replenishing C14 nuclei maintaining a balance a balance. This balance starts getting disrupted as soon as the tree is felled).
If T1/2​ is the half life of C14 then e−× T1/2​ln2​=53​
Hence t=T12​ln2ln5/3​=4105 years =4.1×103 years

Question 19

In a uranium ore the ratio of 238U nuclei to 206 Pb nuclei is h=2.8. Evaluate the age of the ore assuming all the lead 206 Pb to be a final decay product of the uranium series. The half-life of 238 U nuclei is 4.5×109 years. [ ln(14/19)=0.3054]

Accepted Answer

What this implies is that in the time since the ore was formed, 1+ηη​U238 nuclei have remained undecayed. Thus1+ηη​=e−+×T12​ln2​Taking logarithm both side, we get or t=T21​​ln2lnη1+η​​Substituting T21​​=4.5×109 years,η=2.8 we get t=1.98×109 years

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Solution for Class 12, Physics, Nuclei

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