Atomic Nucleus Multiple Choice Questions
Question 1. Suppose we consider a large number of containers each containing initially 10000 atoms of radioactive material with a half-life of 1 year. After 1 year,
- All containers will have 5000 atoms
- All the containers will contain the same number of
- Atoms but that number will only be approximately 5000
- The containers will in general have different numbers of atoms but their average will be close to 5000
- None ofthe containers can have more than 5000 atoms
Answer: 3. The containers will in general have different numbers of atoms but their average will be close to 5000
Question 2. When a nucleus in an atom undergoes radioactive decay, the electronic energy levels of the atom
- Do not change for any type of radioactivity
- Change for α and β -radioactivity but not for y radioactivity
- Change for α -radioactivity but not for others
- Change for β -radioactivity but not for other
Answer: 2. Change for α and β -radioactivity but not for y radioactivity
Question 3. M_{x} and M_{y} denote the atomic masses of the parent and; the daughter nuclei respectively in a radioactive decay. The Q -value for β -decay is Q_{1} and that for a β^{+} -decay is Q_{2}. If m_{e} denote the mass of an electron, then which of the following statements is correct?
- Q_{1} = (M_{x} -M_{y})c² and Q_{2} = (M_{x}-M_{y}-2m_{e})c²
- Q_{1} = (M_{x} – M_{y})c² and Q_{2} = (M_{x}-M_{y})c²
- Q_{1 }= (M_{x}-M_{x}-2m_{e})c² and Q_{2} = (M_{x}-M_{y} + 2m_{e})c²
- Q_{1 } = (M_{x}-M_{x} + 2m_{e})c² and Q_{2} = (M_{x}-M_{y}+ 2m_{e})c²
Answer: 1. Q_{1} = (M_{x} -M_{y})c² and Q_{2} = (M_{x}-M_{y}-2m_{e})c²
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Question 4. Heavy and stable nuclei have more neutrons than protons. This is because the factor
- Neutrons are heavier than protons
- Electrostatic force between protons is repulsive
- Neutrons decay into protons through fi -decay
- Nuclear forces between neutrons are weaker than those between protons
Answer: 2. Electrostatic force between protons are repulsive
Question 5. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used light nuclei. Heavy nuclei will not serve the purpose because,
- They will break up
- Elastic collision of neutrons with heavy nuclei will not slow them down
- The net weight of the reactor would be unbearably high
- Substances with heavy nuclei do not occur in liquid or gaseous state at room temperature
Answer: 2. Elastic collision of neutrons with heavy nuclei will not slow them down
Question 6. Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact
- Nuclear forces have short-range
- Nuclei are positively charged
- The original nuclei must be completely ionized before fusion can take place
- The original nuclei must break up before combining
Answer: 1 and 2
Question 7. The density of the uranium nucleus is approximately
- 10^{20} kg. m^{-3}
- 10^{17} kg. m^{–}^{3}
- 10^{13} kg. m^{–}^{3}
- 10^{11} kg. m^{–} ^{3 }
Which is the correct option? Given, m_{p} – 1.67 × 10^{-27} kg
Answer: 2. 1 0^{17} kg. m^{–}^{3}
The density of all nuclei is almost the same and its magnitude, ρ ≈ 10^{14} g cm^{–} ^{3} = 10^{17 }kg m^{–} ^{3}
Question 8. The approximate value of the density of the uranium nucleus (m_{p} = 1.67 × 10^{27 }kg) is
- 10^{20 }kg. m^{-3}
- 10^{17 }kg. m^{-3}
- 10^{14 }kg. m^{-3}
- 10^{11}kg.m^{-3}
Answer: 2. 10^{17 }kg. m^{-3}
Question 9. Which of the following is correct?
- The rest mass of a stable nucleus is less than the sum of the rest masses of the isolated nucleons
- The rest mass of a stable nucleus is more than the sum of the rest masses of the isolated nucleons
- In nuclear fusion, energy is emitted due to the combination of two nuclei of comparable masses ( lOOu approx)
- In nuclear fission, no energy is released due to fragmentation of a very heavy nucleus
Answer: 1. The rest mass of a stable nucleus is less than the sum of the rest masses of the isolated nucleons
Question 10. During the emission of a negative β -particle
- An electron from the atom is emitted
- An electron already present inside the nucleus is emitted
- An electron is emitted due to the disintegration of a neutron inside the nucleus
- A part of nuclear binding energy is converted into an electron.
Answer: 3. An electron is emitted due to the disintegration of a neutron inside the nucleus
Question 11. Which of the following statements is correct?
- β-rays and cathode rays are identical.
- γ -rays are a stream of highly energetic neutrons
- α -particles are singly charged helium atoms
- The mass of a proton and that of a neutron are exactly equal
Answer: 1. β-rays and cathode rays are identical.
Question 12. A radioactive nucleus of mass number A, initially at rest, emits an a -particle with a speed v. The recoil speed of the daughter nucleus will be
- \(\frac{2 v}{A-4}\)
- \(\frac{2 v}{A+4}\)
- \(\frac{4 v}{A-4}\)
- \(\frac{4 v}{A+4}\)
Answer: 3. \(\frac{4 v}{A-4}\)
Question 13. An excited Ne²² nucleus is disintegrated into an unknown nucleus and two α -particles. This unknown nucleus is
- Nitrogen
- Carbon
- Boron
- Oxygen
Answer: 2. Carbon
Question 14. The half-life of radioiodine I¹³¹ is 8 d. If a sample of I¹³¹ is taken at time t = 0 then it can be said that
- No nuclear disintegration will occur before t = 4 d
- No nuclear disintegration will occur before t 8 d
- All nuclei will be disintegrated in t = 16 d
- A definite nucleus may be disintegrated at any time after t = 0
Answer: 4. A definite nucleus may be disintegrated at any time after t = 0
Question 15. In a freshly prepared radioactive sample, the rate of radia¬ tion is 64 times greater than the safe limit. If its half-life is 2 h then using these safety experiments can be performed safely after
- 6h
- 12 h
- 24 h
- 128 h
Answer: 2. 12 h
Question 16. The mean life of a radioactive element is 13 days. Initially, a sample contains 1 g of this element. The mass of the ele¬ ment will be 0.5 g after a time of
- 13 days
- 9 days
- 18.75 days
- 6.5 days
Answer: 2. 9 days
Question 17. The half-life of At^{215} is 100 μs. The time taken for the radioactivity of the sample ofthe element to decay 1/16 th of its initial value is
- 400 μs
- 6.3μs
- 40 μs
- 300 μs
Answer: 1. 400 μs
Question 18. The half-life of a radioactive substance is 20 min. The approximate time interval (t_{2}– t_{1}) between the time t_{2} when 2/3 of it has decayed and time t_{1} when1/3 of it had decayed is
- 14 min
- 20 min
- 28 min
- 7 min
Answer: 2. 20 min
Hint: Given, half-life T = \(\frac{\ln 2}{\lambda}\)
T = \(\frac{\ln 2}{\lambda}\) = 20 min
We know that, N = \(N_0 e^{-\lambda t}\)
∴ \(\left(1-\frac{2}{3}\right) N_0=N_0 e^{-\lambda t_2}\)
Or, \(\frac{1}{3} N_0=N_0 e^{-\lambda t_2}\)……….(1)
Again \(\left(1-\frac{1}{3}\right) N_0=N_0 e^{-\lambda t_1}\)
Or, \(\frac{2}{3} N_0=N_0 e^{-\lambda t_1}\) …………….(2)
Dividing equation (l) by equation (2), we get
⇒ \(\frac{1}{2}=e^{-\lambda\left(t_2-t_1\right)}\)
∴ \(t_2-t_1=\frac{\ln 2}{\lambda}\)
= 20 min
Question 19. The half-life of radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1: 15 in a sample of a given rock. The age of the rock was estimated to be
- 200 years
- 150 years
- 250 years
- 100 years
Answer: 2. 150 years
Question 20. A nucleus emits one particle and two γ- particles. The resulting, nucleus is
- _{n-4 }Z ^{m-6}
- _{n }Z^{m-6}
- _{n X }^{m-4}
- _{n-2 Y }^{m-4}
Answer: 3. _{n X}^{m-4}
Question 21. Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R . At time t – 0, the number of P species is 4N_{0} and that of Q is N_{0}. The half-life of P (for conversion to R ) is1 min whereas that of Q is 2 min. Initially version of R ) is 1 min whereas that of Q is 2 min. Initially version of R ) is 1 min whereas that of Q is 2 min. Initially R present in the sample would be
- 2N_{0}
- 3N_{0}
- \(\frac{9 N_0}{2}\)
- \(\frac{5 N_0}{2}\)
Answer: 3. \(\frac{9 N_0}{2}\)
Hint: At’ t = 0, the number of nuclei of P and Q respectively 4N_{0} and N_{0}
Let at t = t, the number of nuclei of P and Q are respectively N_{P} and N_{Q}
⇒ \(4 N_0\left(\frac{1}{2}\right)^{t / 1}\)
⇒ \(N_Q=N_0\left(\frac{1}{2}\right)^{t / 2}\)
∴N_{P} = N_{Q}
∴ \(4 N_0\left(\frac{1}{2}\right)^t=N_0\left(\frac{1}{2}\right)^{t / 2}\)
So, \(\frac{4}{(2)^t}=\frac{1}{2^{t / 2}}\)
t = 4 min
∴ After 4 minutes number of atoms of both types is the same; the Number of atoms after 4 minutes,
⇒ \(\left(4 N_0-\frac{N_0}{4}\right)+\left(N_0-\frac{N_0}{4}\right)=\frac{9 N_0}{2}\)
Question 22. Of the following equations which one is the probable nuclear fusion reaction?
- _{ 5}C^{13}+ _{1}H^{1} → _{6}C^{14}+ 4.3 MeV
- _{6}C^{12} + _{1}H^{1} → _{7}N^{13} + 2 MeV
- _{7}N^{14}+ _{1}H^{1 }→ _{8}O^{15}
- _{92}U^{235} + _{0}n^{1 }→ _{34 }Xe^{140 }+ _{38}Sr^{94} + _{0}n^{1 }+ ϒ + 200.Mey
Answer: 2._{6}C^{12} + _{6}C^{12} + _{1}H^{1} → _{7}N^{13} + 2 MeV→ _{7}N^{13} + 2 MeV
Question 23. In the nuclear reaction \({ }_7^{14} \mathrm{~N}+X \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1^1 \mathrm{H}\) X will be
- _{1}H^{1}
- _{1}H^{1}
- _{1}H^{2}
- _{0}n^{1}
Answer: 4. _{0}n^{1}
Question 24. Fast-moving neutrons are retarded
- By using lead obstacle
- By passing through water
- After colliding elastically with heavy nuclei
- Strong electric fields
Answer: 2. By passing through water
Question 25. In nuclear fusion
- A heavy nucleus breaks into two intermediate nuclei and a few high particles
- A light nucleus breaks due to collision with a thermal neutron
- A heavy nucleus breaks due to collision with a thermal neutron
- Two or more light nuclei combine into a heavier nucleus and a few light particles
Answer: 4. Two or more light nuclei combine into a heavier nucleus and a few light particles
Question 26. 4_{1}H^{1 }→ _{2}He^{4} + 2e^{+} + 26 MeV: this is an equation of
- β-decay
- γ-decay
- Fusion
- Fission
Answer: 3. Fusion
Question 27. The power obtained in a reactor using U^{235} disintegration is 1000 kW. The mass decay of U – 235 per hour is
- 10μg
- 20μg
- 40μg
- 1μg
Answer: 3. 40μg
Question 28. A radioactive isotope XA becomes YA~4 after decay. Which ofthe following radioactive emissions are not possible in this case? ‘
- α
- β
- Meson
- Positron
Answer: 2,3 and 4
Question 29. A radioactive isotope X^{A} becomes Y^{A-4} after disintegration. Which ofthe following radioactive emissions are not possible in this case?
- α
- β
- Meson
- Positron
Answer: 1 and 3
Question 30. In the case of a radioactive element which of the following relations are correct where λ = decay constant, T = half-life, and τ = mean life?
- \(\tau=\frac{1}{\lambda}\)
- \(\tau=\frac{0.693}{\lambda}\)
- \(0.6 T\)
- \(\tau=\frac{T}{0.693}\)
Answer: 1 and 4
Question 31. When α -rays and β -rays are compared as radioactive radiation, it is found that ‘
- The deflection of β -particles in an electric or magnetic field is comparatively larger
- The penetration power of β -particles is more
- The ionization power of β -particles Is more
- The velocity of β -particles is more
Answer: 1, 2 and 4
Question 32. The ratio of the mass number of two nuclei is 1: 8, then
- Ratio of diameter =1:4
- Ratio of diameter =1:2
- Ratio of volume =1:8
- Ratio of volume =1:4
Answer: 2, 3
Question 33. In any nuclear reaction
- The total number of protons and neutrons remains the same before and after the reaction
- An increase or decrease in the number of protons is equal to a decrease or increase in the number of neutrons
- Kinetic energy of the incident particle is approximately 8 MeV or its equivalent
- Some energy is released if total mass is reduced
Answer: 1, 3 and 4
Question 34. The initial number of radioactive atoms in a radioactive sample is N_{0}. If after time t the number becomes N, then N = N_{0} e^{-λt}, where λ is known as the decay constant ofthe element. The time in which the number of radioactive atoms becomes half of its initial number is called the half-life (T) of the element. The time in which the number of atoms falls to 1/e times its initial number is the mean life (τ) ofthe element. The product λN is the activity (A) ofthe radioactive sample when the number of atoms is N. The SI unit of activity is bequerel (Bq); where lBq = 1decay. s^{-1}, and Avogadro’s number, N = 6.023 × 10^{23}.
1. The half-life of iodine-131 is 8d. Its decay constant (in SI)
- 10^{-6}
- 1.45 × 10^{-6}
- 2 × 10^{-6}
- 2.9 × 10^{-6}
Answer: 1. 10^{-6}
2. The half-life of iodine-131 is 8d. Its mean life (in SI ) is
- 4.79 × 10^{5}
- 6.912× 10^{5}
- 9.974× 10^{5}
- 22. 96 × 10^{5}
Answer: 3. 9.974× 10^{5}
3. The half-life of Iodine 131 is 8d. What is the activity Bq) of 1 g of iodine?
- 2.3 × 10^{15}
- 4.6 × 10^{15}
- 6.9 × 10^{15}
- 9.2 × 10^{15}
Answer: 2. 4.6 × 10^{15}
4. in the equation above After how many days the activity of iodine-131 will be \(\frac{1}{16}\)th of its initial value
- 24 data
- 32 data
- 40 data
- 48 data
Answer: 2. 32 data
5. In the question above, what is the ratio of the activity of sodium- 24 to that of iodine-131 (half-life of sodium- 24 is 15h)?
- \(\frac{1}{70}\)
- \(\frac{1}{7}\)
- 7
- 70
Answer: 4. 70
Question 35. For the radioactive nuclei that undergo either a or /S decay, which one of the following cannot occur?
- Isobar of the original nucleus is produced
- Isotope of the original nucleus is produced
- Nuclei with higher atomic numbers than that of the original nucleus is produced
- Nuclei with lower atomic number than that of the
Answer: 2. Isotope of the original nucleus is produced
Question 36. Radon-222 has a half-life of 3.8 days. If one starts with 0.064 kg of radon-222 the quantity of radon-222 left after 19 days
will be
- 0.002 kg
- 0.032 kg
- 0.062 kg
- 0.024 kg
Answer: 1. 0.002 kg
⇒ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^{t / T}\)
Or, \(N=N_0\left(\frac{1}{2}\right)^{19 / 3.8}\)
= \(0.064 \times \frac{1}{32}\)
= 0.002 kg
Question 37. If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the 3rd day? (given, \(\sqrt[3]{0.25}\) = 0.63 )
- 0.63
- 0.37
- 0.5
- 0.13
Answer: 4. 0.13
We know, in case of radioactive decay
N = \(N_0 e^{-\lambda t}\)
Again = \(\lambda=\frac{\ln 2}{T_{1 / 2}}\)
Given T = 3 day
∴ \(\lambda=\frac{\ln 2}{3}\)
The fraction ofthe initial number of nuclei will decay on the 3rd day
= \(\frac{N_0 e^{-\frac{\ln 2}{3} \times 2}-N_0 e^{-\frac{\ln 2}{3} \times 3}}{N_0}=e^{-\frac{2 \ln 2}{3}}-e^{-\frac{3 \ln 2}{3}}\)
= \(2^{-\frac{2}{3}}-2^{-1}\)
= 0.13
Question 38. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have an equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be
- 1:16
- 4:1
- 1:4
- 5:4
Answer: 4. 5:4
80 minutes = 4 half-lives of A = 2 half-lives of B
Let the initial number of nuclei in each sample be N.
Number of undecayed nuclides of element A after 80 minutes \(\)
Number of A nuclides decayed = \(\frac{N}{2^2}\)
Number of undecayed nuclides of element B after 80 minutes = \(\frac{N}{2^2}\)
Number of B nuclides decayed \(\frac{3}{4}\)
Required ratio = \(\frac{15 / 16}{3 / 4}=\frac{5}{4}\)
Question 39. A radioactive nucleus A with a half-life of T, decays into a nucleus B. At t = 0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by
- t = \(\frac{T}{2} \frac{\log 2}{\log 1.3}\)
- t = \(T \frac{\log 1.3}{\log 2}\)
- t = \(T \log (1.3)\)
- t = \(\frac{T}{\log (1.3)}\)
Answer: 2. t = \(T \frac{\log 1.3}{\log 2}\)
After time t, the number of nuclei of A
⇒ \(N_B=N_0-N_A=N_0\left(1-e^{-\lambda t}\right)\)
⇒ \(\frac{N_B}{N_A}\) = 0.3
⇒ \(\frac{1-e^{-\lambda t}}{e^{-\lambda t}}\) = 0.3
t = In 1.3
\(\left(\frac{\ln 2}{T}\right) t\)= In 1.3
t = \(T \frac{\ln 1.3}{\ln 2}=T \frac{\log 1.3}{\log 2}\)
Question 40. The binding energy per nucleon of _{3}Li^{7 }and _{2}He^{4 } nuclei are 5.60MeV and 7.06MeV respectively. In the nuclear reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{Q}\) the value of energy Q released is
- 19.6 MeV
- – 2.4 meV
- 8.4 MeV
- 17.3 MeV
Answer: 4. 17.3 MeV
Binding energy of _{2}He^{4} = 4 ×7.06 MeV
Binding energy of _{3}Li^{7} = 7 × 5.60 MeV
The nuclear equation is
⇒ \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{Q}\)
7 × 5.60 = 4 × 7.06 + 4 × 7.06 + Q
or, Q = 56.48- 39.20 = 17.28 MeV
Question 41. A radioisotope X with a half-life of 1.4 × 10^{9} years decays to stable Y. A sample of the rock from a cave was found to contain X and Y in a ratio of 1:7. The age of the rock is
- 1.96 × 10^{9} years
- 3.92 × 10^{9} y
- 3.92 × 10^{9} y
- 4.20 × 10^{9} years
Answer: 3. 3.92 × 10^{9} y
Suppose N atoms out of N_{0} atoms of element X disintegrate to element Y in time t
∴ \(\frac{N}{N_0-N}=\frac{1}{7} \quad \text { or, } \frac{N}{N_0}=\frac{1}{8}\)
∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^3\)
∴ Time taken for this disintegration is 3 times the half-life.
t = 3 × 1.4 × 10^{9} = 4.2 × 10^{9} years
Question 42. If the radius of the _{13}Li^{27} nucleus is taken to be RM, then the 125 radius of the _{53}Te^{125} nucleus is nearly
- \(\left(\frac{53}{13}\right)^{1 / 3} R_{\mathrm{Al}}\)
- \(\frac{5}{3} R_{\mathrm{Al}}\)
- \(\frac{3}{5} R_{\mathrm{Al}}\)
- \(\left(\frac{13}{53}\right)^{1 / 3} R_{\mathrm{Al}}\)
Answer: 2. \(\frac{5}{3} R_{\mathrm{Al}}\)
We know R = \(r_0 \mathrm{~A}^{1 / 3}\)
Then R Al = \(r_0(27)^{1 / 3}\) = 3rd
= \(R_{\mathrm{Te}}=r_0(125)^{1 / 3}\) = 5r
= \(\frac{5}{3} \cdot 3 r_0=\frac{5}{3} R_{\mathrm{Al}}\)
Question 43. The energy liberated per nuclear fission 10zo fissions occur per second the amount of power produced will be
- 2 × 10^{22} W
- 32 × 10^{8} W
- 16 × 10^{8} W
- 5 × 10^{11} W
Answer: 2. 32 × 10^{8} W
Power = Energy liberated per nuclear fission x number of fissions per second
= 200 × 10^{20} MeV/s
= (200 × 10^{6} × 1. 6 × 10^{-19}) × 10^{20} J/s
= 3.2 × 10^{9}
= 32 × 10^{8}W
Question 44. For a radioactive material, the half-life is 10 minutes. If initially there are 600 nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
- 3
- 10
- 20
- 15
Answer: 3. 20
⇒ \(\left(\frac{N}{N_0}\right)=\left(\frac{1}{2}\right)^{t / T}\)
⇒ \(\frac{600-450}{600}=\left(\frac{1}{2}\right)^{t / 10}\)
or, t/10 = 2
t = 20 min
Question 45. A radioactive element emits 2 or -particles and 3 0 – particles. The values of atomic number (Z) and mass number (A) of the new element will be
Answer:
- (A +5),(Z-1)
- (A – 5) ,(Z+1)
- (A -8),(Z-1)
- (A – 8) ,(Z+1)
Answer: 3. (A -8),(Z-1)
The mass number after emission of 2 or -particles
= A – (2 × 4) = (A – 8)
The atomic number after emission of 2 α -particles and 3 β particles
Z- (2 × 2) + (3 × 1) = Z-1