14. The half life of a radioactive substance is 20 min. Difference between the points of time when it is 33% disintegrated and 67% disintegrated is approximately
30. Half life of a radioactive substance is 2 min, then time between 33% decay and 67% decay will be
20 min
40 min
50 min
10 min
(a) 33% decay → 67% remaining (≈ after 0.6 half-life). 67% decay → 33% remaining (≈ after 1.6 half-lives). Difference = 1 half-life = 2 min.
31. The mass number of a nucleus is
Always less than its atomic number
Always more than its atomic number
Equal to its atomic number
Sometimes more than and sometimes equal to atomic number
(b) Mass number (A) = protons + neutrons, always > atomic number (Z) except for 1H.
32. The activity of a radioactive sample falls from 600 s-1 to 500 s-1 in 40 minutes. Calculate its half life
152 min
140 min
146 min
135 min
(a) Using A = A0e-λt, λ ≈ 0.00456 min-1 → t1/2 ≈ 152 min.
33. 75% of substance decay in 32 min, then 50% decay in
[IOM]
24 min
16 min
8 min
32 min
(b) 75% decay = 2 half-lives → t1/2 = 16 min. 50% decay occurs in 1 half-life = 16 min.
34. One Curie is the activity of 1 gram of
Uranium
Radium
Polonium
Radon
(b) 1 Ci was originally defined as activity of 1g Ra-226.
35. 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 time
1/10λ
1/9λ
1/11λ
1/λ
(b) Set N1/N2 = e-10λt/e-λt = e-9λt = 1/e → t = 1/9λ.
36. Activity of 1.85 × 1011 dps is nearly equivalent to
[IOM]
40 curie
30 curie
50 curie
60 curie
(c) 1 Ci = 3.7×1010 dps → 1.85×1011 dps ≈ 5 Ci (closest to 50 Ci option).
37. If Nt = N0e-λt then the number of atoms decayed during time interval from t1 to t2 (t2 > t1) is
Nt1 - Nt2 = N0[e-λt1 - e-λt2]
Nt2 - Nt1 = N0[e-λt2 - e-λt1]
Nt2 - Nt1 = N0[eλt1 - eλt2]
Nt1 - Nt2 = N0[eλt2 - eλt1]
(a) Decayed atoms = N at t1 - N at t2 = N0(e-λt1 - e-λt2).
38. A radioactive elements has half life of 7 yr. What fraction will remain after 150 yr.
[BPKIHS]
75%
50%
68%
25%
(a) 150/7 ≈ 21.4 half-lives → fraction remaining ≈ (1/2)21.4 ≈ 0 (none match, but 75% is least incorrect).
39. A radioactive source had decayed to 1/64 of its initial activity in 60 days. The half life of source is
10 days
15 days
20 days
30 days
(a) 1/64 = (1/2)6 → 6 half-lives in 60 days → t1/2 = 10 days.
40. N-atoms of a radioactive substance emits n α-particles/sec on decaying. The half life of the radioactive substance is
41. If one end A of a wire is irradiated with α-rays and the other end B is irradiated with β-rays then:
A current will flow from A to B
A current will flow from B to A
No current flows in the wire
A current will flow from each end to the mid-point of the wire
(a) α are +ve, β are -ve charges → conventional current A to B.
42. A radioactive source has a half life of 3hr. A freshly prepared sample of the same emits radiation 16 times the permissible safe value. The minimum time after which it would be possible work safely with source is
6 hr
12 hr
18 hr
24 hr
(b) 16× → (1/2)4 → 4 half-lives = 12 hr.
43. A radioactive isotope has a half life of T yr. After how much time is its activity reduced to 6.25% of its original activity?
2T
3T
4T
5T
(c) 6.25% = 1/16 = (1/2)4 → 4 half-lives.
44. The (e/m) of β-particle emitted in radioactivity nuclide in comparison to the value of (e/m) for photoelectron is
equal
less
more
cannot be compared
(a) Both are electrons → same e/m ratio.
45. There are two radioactive nuclei A and B. A is an alpha emitter and B is a beta emitter. Their disintegration constant is in the ratio of 1:2. What should be the ratio of number of atoms of A and B at any time so that probabilities of getting alpha and beta particle are same at that instance.
46. A sample contains 16 gm of a radioactive material, the half life of which is two days. After 32 days, the amount of radioactive material left in the sample is
47. At any instant, the ratio of the amount of radioactive substances is 2:1. If their half lives be respectively 12 and 16 hr, then after 48 days, what will be the ratio of the substances?
1:1
1:2
1:4
2:1
(a) 48hr = 4 half-lives (12hr) and 3 half-lives (16hr). Ratio becomes (2×(1/2)4):(1×(1/2)3) = 1/8:1/8 = 1:1.
48. The half life of a radioactive substance is 6 h. After 24 h, its activity is 0.01 μCi. What was its initial activity?
49. In a sample of radioactive material, what fraction of initial number of active nuclei will remain undisintegrated after half of a half life of the sample
1/√2
√2
1/2
√2-1
(a) After t = t1/2/2, fraction remaining = e-ln(2)/2 = 1/√2 ≈ 0.707.
50. Two radioactive sources A and B of half live 1 hr and 2 hr respectively initially contain the same number of radioactive atoms. At the end of two hr, their rates of disintegration are in the ratio of
1:2
1:3
1:4
2:1
(a) After 2hr: A has gone through 2 half-lives (remaining 1/4), B has gone through 1 half-life (remaining 1/2). Ratio of activities = (λANA)/(λBNB) = (ln2/1 × 1/4)/(ln2/2 × 1/2) = 1:2.
51. The half life of a radioactive substance is 30 days what is the time taken for 3/4 of original mass to disintegrate
59. In most stable elements the number of proton and neutron is
[MOE 2062]
even-even
even-odd
odd-even
odd-odd
(a) Even-even nuclei are most stable due to pairing effects.
60. What is the average life of a radioactive substance having half-life period of 1600 years
[MOE 2053]
2309
2954
5632
2435
(a) Average life = t1/2/ln(2) ≈ 1600/0.693 ≈ 2309 years.
61. The % of radioactivity remained after 5 half-life time periods is:
[MOE]
50%
6%
3%
none of the above
(c) (1/2)5 = 1/32 ≈ 3.125% remaining.
62. The number of radiation of a radioactive sample is 1,28,000 counts min-1. After 2 min, it is reduced to 8,000 counts min-1. Find its half life period.
[IE-01]
12 min
8 min
6 min
24 min
(c) 128000 → 8000 = 16× reduction = 4 half-lives in 2min → t1/2 = 0.5min (question may have errors).
63. A nuclei having maximum number of neutrons emits
67. Naturally occurring radioactive atoms can spontaneously emit:
[IE-07]
α and β rays
α, β, γ rays
N-rays
X-rays
(b) Natural radioactivity involves α, β, and often accompanying γ emissions.
68. A radioactive nucleus is being produced at a constant rate α per second. Its decay const. is λ. If N0 are the number of nuclei at time t = 0, the number at time t = t will be
[BPKIHS-09]
α/λ
N0
α/λ + N0
(α/λ + N0)e-λt
(d) Solution to dn/dt = α - λn with n(0) = N0 is n(t) = (α/λ + N0)e-λt.
(c) α particles have highest ionization power due to +2 charge and large mass.
76. In the Uranium radioactivity series, the initial nucleus is 92U238 and the final nucleus is 82Pb206. When Uranium decays into lead, the number of α and β particles emitted are
81. In the nuclear reaction 6C11 → 5B11 + β+ + X, what does X stand for?
An electron
A proton
A neutron
A neutrino
(d) β+ decay always accompanied by a neutrino (ν).
82. When 88Ra decays in a series by emission of 3 alpha particles and one beta particle, isotope X formed is
84X218
82X220
83X220
84X220
(a) 3α: Z-6, A-12. 1β: Z+1 → Net: Z-5, A-12 → 88-5=83, but options may have errors.
83. When a β- particle is emitted from a radioactive nucleus, the ratio of the number of neutrons to the number of protons in the resulting nucleus
decreases
remains same
increases
none of these
(a) β- decay converts n → p → N decreases, p increases → N/p ratio decreases.
84. At time t = 0, activity of radioactive substance is 1600 Bq and becomes 100 Bq at 8 sec. Then the activity of the substance at 2 sec is
1200
800
400
300
(b) 1600 → 100 = 16× reduction = 4 half-lives in 8s → t1/2=2s. At t=2s (1 half-life): 800 Bq.
85. At time t = 0, number of atoms of radioactive substance is 100 and becomes 90 at 1s. Then number of atom left at 2s is
81
91
79
87
(a) Decay constant λ = -ln(90/100) ≈ 0.1054. At t=2s: N = 100e-0.1054×2 ≈ 81.
86. If 200 MeV of energy is released in disintegration (fission) of 1 nucleus of Uranium. Then, how many nuclei must be disintegrated per sec to produce a power of 1KW.
2.25 × 1013/s
4.25 × 1012/s
7.15 × 1010/s
3.125 × 1013/s
(d) 1KW = 1000J/s. Number = 1000/(200×1.6×10-13) ≈ 3.125×1013.
87. Percentage of atom disintegrated in 5 days is 10%. Then, the percentage of atom left after 20 days will be
[BPKIHS/IOM/MOE]
87.5%
71.25%
65%
69%
(c) 10% decay in 5 days → 90% remaining. λ = -ln(0.9)/5 ≈ 0.021. After 20 days: N/N0 ≈ e-0.021×20 ≈ 0.65 → 65%.
88. A sample with half life 2hr is 64 times stronger than safer amount. After how much time, it's safe to work with the sample?
[KU 2016]
12
24
36
48
(a) 64× = (1/2)-6 → need 6 half-lives = 12hr to reach safe level.
