Radioactivity is the phenomenon exhibited by the atomic nucleus due to its instability. The unstable nucleus will emit radiation and energy is lost due to the radiation emitted by the nucleus. The radioactive nucleus will emit energetic particles such as photons, neutrinos, electrons, neutrons, protons or alpha particles. If the nucleus of the atom has too many neutrons, it will emit beta particles and one of the neutrons changes into a proton. If there are too many protons in a nucleus, it will emit a positron changing a proton into a neutron. The nucleus emitting gamma rays will give out a lot of energy without changing any particle of the nucleus. If the nucleus has too much mass, it will emit alpha particles, discarding two neutrons and two protons.
Download JEE Advanced Previous Year Questions with Solutions on Radioactivity PDF
Question 1)To determine the half life of a radioactive element, a student plots a graph of
Answer: (8)
Solution:
(dN/dt) ∝ – N
(dN/dt) = – λ (N0e-λt)
Taking log on both sides
Slope = λ = (4-3)/(6-4)
= ½
⇒t1/2 = 0.693/ λ
= 0.693 x 2 = 1.48 year
Since the number of radioactive nuclei of the element decreases by the value p
N = N0 (½)n
= N0(½)t/t1/2
= N0(½)4.46/1.486
⇒ (N/N0) = (½)(4.46/1.486)
(N/N0) = (½)3
(N/N0) = ⅛ = 1/p
⇒ p = 8
Question 2) The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10-25kg. The mass (in mg) of the radioactive sample is
Answer: 1mg
Solution:
Given
(dN/dt) = 1010
Mean Life = 109 s
⇒ (1/λ) = 109 s
λ = 10-9 s-1
First order decay :
(dN/dt) = λN
λN = 1010
N = 1010/10-9
= 1019
Mass of the radioactive sample = 10-25 x 1019
= 10-6 kg
= 1 mg
Question 3) The β-decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e–) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process,
If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?
(A) 0 ≤ K ≤ 0.8 × 106 eV
(B) 3.0 eV ≤ K ≤ 0.8 × 106 eV
(C) 3.0 eV ≤ K ≤ 0.8 × 106 eV
(D) 0 ≤ K< 0.8 × 106 eV
Answer: (D) 0 ≤ K< 0.8 x 106 eV
Solution:
K will not be equal to 0.8 × 106 eV as anti-neutrino must have some energy.
Question 4) A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is.
Answer: 4%
Solution:
Given: t1/2 = 1386 s
λ = ln2/t1/2
= 0.693/1386
= 5 × 10-4 s-1
t = 80 s
N = N0e-λt
N0 – N = number of nuclei that are decayed
1 – (N/N0) = 1 – e-λt
(N0 – N)/N0 = 1 – e-λt
[(N0 – N)/N] x 100 = (1 – e-λt) × 100= (1 – e (-0.000 5 x 80)) × 100
= (1 – e – 0.04) × 100
If x＜＜1 then e-x = (1- x)
= [1 – (1 – x)] × 100
= 1- (1 – 0.04) × 100
= 4%
Question 5) A nuclear power plant supplying electrical power to a village uses a radioactive material of half-life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is.
Answer: 3
Solution:
12.5% is (⅛) th of the initial power
After each T, P will become half
Therefore, total time = 3T
n = 3
Question 6) For radioactive material, its activity A and rate of change of its activity R are defined as A= – (dN/dt) and R= (-dA/dt), where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life τ) and Q (mean life 2τ) have the same activity at t = 0. Their rates of change of activities at t = 2τ are Rp and RQ, respectively. If Rp/RQ = n/e, then the value of n is:
Answer: 2
Solution:
N = N0 e-λt
A = (-dN/dt)
R = – dA/dt
= d2N/dt2
= N0 λ2 e-λt
=λ (λN0) e-λt
=λ A0 e-λt
On comparing this with Rp/RQ = n/e,
n = 2
Question 7) An accident in a nuclear laboratory resulted in the deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for the safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
(a) 64
(b) 90
(c) 108
(d) 120
Answer: (C) 108
Solution:
T1/2 = 18 days
R = 64Rpermissible
Using
t = (2.303 × 1.806 × 18)/0.693
= 108 days
Question 8) In a radioactive decay chain, 23290 Th nucleus decays to 21282b nucleus. Let Nα and Nβ be the number of α and β particles, respectively, emitted in this decay process. Which of the following statements is(are) true?
(a) Nα = 5
(b) Nα = 6
(c) Nβ = 2
(d) Nβ = 4
Answer: (A) Nα = 5 (C) Nβ = 2
Solution:
23290 Th nucleus decays to 21282b nucleus
Change in the mass number (A) = 20
Number of α particles = 20/4 = 5
90 – 2Nα + Nβ = 82
90 – 2(5) + Nβ = 82
Nβ = 2
Question 9) Which of the following is a correct statement?
(a) Beta rays are the same as cathode rays
(b) Gamma rays are high energy neutrons
(c) Alpha particles are singly ionised helium atoms
(d) Protons and neutrons have exactly the same mass
Answer: (a) Beta rays are the same as cathode rays
Solution:
Beta rays and cathode rays will have the accelerated electrons
Gamma rays are high energy photons
Alpha particles are doubly ionised helium atoms
Protons and neutrons have different mass
Question 10) A radioactive sample of 238U decays to Pb through a process for which the half-life is 4.5 x 109 years. Find the ratio of the number of nuclei of Pb to 238U after a time of 1.5 x 109 years. (given 21/3 = 1.26)
Answer: 0.26
Solution:
NPb = 0.26Nu
NPb/Nu = 0.26
Question 11) At a given instant there are 25% undecayed radio-active nuclei in a sample. After 10 seconds the number of undecayed nuclei reduces to 12.5%. Calculate (i) the mean life of the nuclei, and (ii) the time in which the undecayed nuclei will further reduce to 6.25% of the reduced number
Answer: 14.43 s, 40 s
Solution:
(i) mean life of the nuclei is
τ = τ1/2/0.693
= 10/0.693
= 14.43 s
(ii) From 12.5% nuclei has been reduced to 6.25%
N = N0(½)n
6.25 = 100 (½)n
2n = 16
n = 4
So, in 4 half-lives, undecayed nuclei will be further reduced to 6.25% of the reduced number
t = 4 × τ1/2
= 4 × 10 = 40 s
Question 12) A freshly prepared radioactive source of half-life 2 hours 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
(a) 6 hr
(b) 12 hr
(c) 24 hr
(d) 128 hr
Answer: (b) 12 hr
Solution:
Half-life of a radioactive substance, T = 2 hours
Amount of radioactive material left after time t
t = N/N0
⇒ t = (½) t/T
(N/N0) = 1/64
(1/64) = (½) t/T
(½)6 = (½) t/T
⇒ t/T = 6
t = 6 × 2 = 12 hours
Question 13) During a negative beta decay
(a) an atomic electron is ejected
(b) an electron that is already present in the nucleus is ejected
(c) a neutron in the nucleus decays emitting an electron
(d) a part of the binding energy of the nucleus is converted into an electron
Answer: (c) a neutron in the nucleus decays emitting an electron
Solution:
The equation for negative β decay is expressed by the equation
So during negative beta decay, the neutron will decay into a proton, a negative beta particle (electron) and an antineutrino.
Question 14) A small quantity of solution containing Na24 radionuclide(half-life = 15 hours) of activity 1.0 microcurie is injected into the blood. A sample of the blood of volume 1cm3 taken after 5h shows an activity of 296 disintegrations per minute. Determine the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (1 curie =3.7 x 1010 disintegrations per second)
Answer: 5.95 L
Solution:
t1/2 = 15 hr
λ = ln 2/t1/2
= 0.693/15
= 0.462 hr-1
Initial activity = 1 mC = 3.7 × 1010 × 10-6 = 3.7 × 104 dis/sec
The activity of blood of volume 1 cm3 after 5 hours = 296 dis/min
= 296/60
= 4.93 dis/sec
Let R be the activity of the whole volume of blood after 5 hours
R = rV
⇒ V = R/r
= R0e-λt/r
= [3.7 × 104 × e (-0.0462 x 5)] / 4.93
= [3.7 × 104 × e (-0.231)] / 4.93
=( 3.7 × 104 × 0.7937)/4.93
= 5.957 × 103 cm3
V = 5.957 L
Question 15) A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which the one-fourth of the material remains is
(a) 1080
(b) 2430
(c) 3240
(d) 4860
Answer: (a) 1080 yr
Solution:
From Rutherford-Soddy law, the number of atoms left after n half-lives is given by
N = N0 (½)n
N0 is the original number of atoms
The number of half-life n = time of decay/effective half-life
Effective disintegration constant, λ = ln 2/T
Therefore, λ1 + λ2 = (ln 2/T1)+ (ln 2/T2)
Effective half life,
= (1/1620) + (1/810)
⇒ T = 540 yr
n = t/540
N = N0 (½)t/540
⇒ t/540 = 2
t = 2 × 540 = 1080 yr
Also Read:
Radioactivity JEE Main Previous Year Questions With Solutions
Nuclei and Radioactivity – Important Topics
Nuclei and Radioactivity – Important Questions
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