Q: In a radioactive decay experiment, the starting number of nuclei is 3000. After 1.0 second, it is observed that 1000 ± 40 nuclei have undergone decay. Using the approximation ln(1 + x) ≈ x for |x| << 1, what is the uncertainty Δλ in calculating the decay constant λ (in s⁻¹)?

0.02. The uncertainty Δλ in calculating the decay constant λ is 0.02, which can be determined by applying the approximation ln(1 + x) ≈ x for small x to the equation describing the radioactive decay and then using the given data to calculate the uncertainty.

Q: A radioactive isotope 40/19 K undergoes decay either into stable 40/20 Ca with a decay constant of 4.5 × 10⁻¹⁰ per year or into stable 40/18 Ar with a decay constant of 0.5 × 10⁻¹⁰ per year. Assuming that all stable 40/20 Ca and 40/18 Ar nuclei originate from the decay of 40/19 K, determin

9.2. The decay constants for 40/19 K allow calculation of the total decay rate. Using the ratio of stable nuclei to remaining nuclei and the decay equation, the time required for the ratio to reach 99 is found to be 9.2 × 10⁹ years.

Q: A radioactive nucleus Q, with a half-life of 20 minutes, decays by emitting alpha particles 60% of the time and beta particles 40% of the time. If the initial count of Q nuclei is 1000, how many nuclei undergo alpha decay within the first hour?

525. One hour equals three 20-minute half-lives, so remaining nuclei = 1000/8 = 125 and decayed = 875. Since 60% of decays are alpha, alpha decays = 0.6 x 875 = 525. The stored answer 350 is wrong; the correct option is 525.

Q: A proton strikes a stationary deuteron to form a He-3 nucleus. For this reaction to occur, the proton must carry a minimum kinetic energy of 1.4 MeV. If the roles are reversed and a deuteron strikes a stationary proton (to form the same He-3 nucleus), what is the minimum kinetic energy the

2.8 MeV. The CM-frame threshold energy is fixed by the Q-value. For a projectile of mass m hitting a target of mass M at rest, E_th = E_cm * (m+M)/(2M). With proton (mass 1u) on deuteron (mass 2u): E_th = E_cm * 3/4. With deuteron (mass 2u) on proton (mass 1u): E_th = E_cm * 3/2. The ratio of the two thresholds is (3/2)/(3/4) = 2, so deuteron threshold = 2 * 1.4 = 2.8 MeV.

Q: An unstable radioactive nucleus X decays into two stable nuclei Y and Z simultaneously (branching decay). At t = 0, only X is present. The decay of X follows first-order kinetics with overall decay constant lambda. Three graphs are plotted: ln(N_X) vs t is a straight line with slope equal

Ratio of N_Y: N_Z at any time is constant.. For branching radioactive decay: dN_X/dt = -(lambda_Y + lambda_Z) * N_X, so overall lambda = lambda_Y + lambda_Z = tan(theta) (magnitude of slope of ln N_X vs t graph). Since dN_Y/dt = lambda_Y * N_X and dN_Z/dt = lambda_Z * N_X, dividing gives dN_Y/dN_Z = lambda_Y/lambda_Z = constant. Integrating, N_Y/N_Z = lambda_Y/lambda_Z = b/c = constant at all times (option 1 correct, option 4 wrong). The partial decay constant lambda_Y = lambda * b/(b+c) = b*tan(theta)/(b+c) (option 2 correct). Option 3 has e^a in the denominator which is incorrect; the correc

Q: Which of the following statements about nuclear physics are correct? (One or more may be correct.)

The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons.. A stable nucleus has a mass defect — its rest mass is less than the sum of individual nucleon masses, with the difference appearing as binding energy. Fusion releases energy for light nuclei (like H isotopes), not medium-mass nuclei. Fission releases energy for very heavy nuclei (like U-235). So options A and D are correct.

Question 1

The correct statement is:

Accepted Answer

Deuteron and alpha particle can undergo complete fusion. Deuteron and alpha particles can undergo complete fusion due to their nuclear properties, which allow them to overcome the Coulomb barrier and form a heavier nucleus under suitable conditions.

Question 2

In the nuclear fission reaction 235⁹²U → 140⁵⁴Xe + 94³⁸Sr + x + y, where x and y represent two particles, the 235⁹²U nucleus is initially stationary. The kinetic energies of the resulting products are denoted as Kₓ, K_Sr, K_Xe (2 MeV), and K_y (2 MeV), respectively. The binding energy per

Accepted Answer

x is a neutron, y is a neutron, K_Sr = 129 MeV, K_Xe = 86 MeV. In nuclear fission, conservation of mass-energy and momentum applies. The binding energy difference indicates two neutrons are emitted (x and y). The kinetic energy distribution aligns with K_Sr = 129 MeV and K_Xe = 86 MeV, satisfying energy conservation.

Question 3

An accident in a nuclear laboratory resulted in 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 safe operation of the laboratory. What is the minim

Accepted Answer

108. The radiation decreases by half every 18 days. Starting at 64 times the permissible level, it takes 6 half-lives (18 × 6 = 108 days) for the radiation to drop to the safe level.

Question 4

The electrostatic potential energy of Z protons uniformly distributed within a spherical nucleus of radius R is expressed as E = (3/5) * Z(Z−1)e² / (4πε₀R). The masses of neutron, ¹H, ¹⁵₇N, and ¹⁵₈O are 1.008665 u, 1.007825 u, 15.000109 u, and 15.003065 u, respectively. If the radii of ¹⁵₇

Accepted Answer

3.42 fm. The binding energy difference between ¹⁵₇N and ¹⁵₈O is attributed to the electrostatic potential energy difference. Using the given formula and data, the radius of the nucleus is calculated to be 3.42 fm.

Question 5

In a radioactive decay experiment, the starting number of nuclei is 3000. After 1.0 second, it is observed that 1000 ± 40 nuclei have undergone decay. Using the approximation ln(1 + x) ≈ x for |x| << 1, what is the uncertainty Δλ in calculating the decay constant λ (in s⁻¹)?

Accepted Answer

0.02. The uncertainty Δλ in calculating the decay constant λ is 0.02, which can be determined by applying the approximation ln(1 + x) ≈ x for small x to the equation describing the radioactive decay and then using the given data to calculate the uncertainty.

Question 6

A radioactive isotope 40/19 K undergoes decay either into stable 40/20 Ca with a decay constant of 4.5 × 10⁻¹⁰ per year or into stable 40/18 Ar with a decay constant of 0.5 × 10⁻¹⁰ per year. Assuming that all stable 40/20 Ca and 40/18 Ar nuclei originate from the decay of 40/19 K, determin

Accepted Answer

9.2. The decay constants for 40/19 K allow calculation of the total decay rate. Using the ratio of stable nuclei to remaining nuclei and the decay equation, the time required for the ratio to reach 99 is found to be 9.2 × 10⁹ years.

Question 7

A radioactive nucleus Q, with a half-life of 20 minutes, decays by emitting alpha particles 60% of the time and beta particles 40% of the time. If the initial count of Q nuclei is 1000, how many nuclei undergo alpha decay within the first hour?

Accepted Answer

525. One hour equals three 20-minute half-lives, so remaining nuclei = 1000/8 = 125 and decayed = 875. Since 60% of decays are alpha, alpha decays = 0.6 x 875 = 525. The stored answer 350 is wrong; the correct option is 525.

Question 8

A proton strikes a stationary deuteron to form a He-3 nucleus. For this reaction to occur, the proton must carry a minimum kinetic energy of 1.4 MeV. If the roles are reversed and a deuteron strikes a stationary proton (to form the same He-3 nucleus), what is the minimum kinetic energy the

Accepted Answer

2.8 MeV. The CM-frame threshold energy is fixed by the Q-value. For a projectile of mass m hitting a target of mass M at rest, E_th = E_cm * (m+M)/(2M). With proton (mass 1u) on deuteron (mass 2u): E_th = E_cm * 3/4. With deuteron (mass 2u) on proton (mass 1u): E_th = E_cm * 3/2. The ratio of the two thresholds is (3/2)/(3/4) = 2, so deuteron threshold = 2 * 1.4 = 2.8 MeV.

Question 9

An unstable radioactive nucleus X decays into two stable nuclei Y and Z simultaneously (branching decay). At t = 0, only X is present. The decay of X follows first-order kinetics with overall decay constant lambda. Three graphs are plotted: ln(N_X) vs t is a straight line with slope equal

Accepted Answer

Ratio of N_Y: N_Z at any time is constant.. For branching radioactive decay: dN_X/dt = -(lambda_Y + lambda_Z) * N_X, so overall lambda = lambda_Y + lambda_Z = tan(theta) (magnitude of slope of ln N_X vs t graph). Since dN_Y/dt = lambda_Y * N_X and dN_Z/dt = lambda_Z * N_X, dividing gives dN_Y/dN_Z = lambda_Y/lambda_Z = constant. Integrating, N_Y/N_Z = lambda_Y/lambda_Z = b/c = constant at all times (option 1 correct, option 4 wrong). The partial decay constant lambda_Y = lambda * b/(b+c) = b*tan(theta)/(b+c) (option 2 correct). Option 3 has e^a in the denominator which is incorrect; the correc

Question 10

Which of the following statements about nuclear physics are correct? (One or more may be correct.)

Accepted Answer

The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons.. A stable nucleus has a mass defect — its rest mass is less than the sum of individual nucleon masses, with the difference appearing as binding energy. Fusion releases energy for light nuclei (like H isotopes), not medium-mass nuclei. Fission releases energy for very heavy nuclei (like U-235). So options A and D are correct.

Question 11

In beta-minus decay, the Q-value of the process is E. Which of the following statements are correct?

Accepted Answer

The kinetic energy of the emitted beta particle cannot exceed E.. In beta-minus decay, a neutron converts to a proton, decreasing the N/Z ratio. The Q-value E is shared between the beta particle and the antineutrino (recoil is negligible), so neither can individually exceed E. Thus statements A and C are correct; B is also correct since the antineutrino KE ranges from 0 to E.

Question 12

In a nuclear fission reaction, U-236 splits as: U-236(92) -> X-117(53) + Y-117(53) + n + n. The binding energy per nucleon of X and Y is 8.5 MeV, while that of U-236 is 7.6 MeV. If the total energy liberated equals 25N MeV, find the integer value of N.

Accepted Answer

N = 8. Energy released = 2*117*8.5 - 236*7.6 = 1989 - 1793.6 = 195.4 MeV. With 25N = 195.4, N ~ 7.8. The problem states 7.58 MeV for U-236 in the original, giving 2*117*8.5 - 236*7.58 = 1989 - 1788.88 = 200.12 MeV, so 25N ~ 200, N = 8.

Question 13

Two radioactive samples A1 and A2 have half-lives of 3 years and 2 years respectively. Both have been decaying for a long time. Today the number of atoms in sample A1 is twice the number of atoms in sample A2. Find the number of years X that have elapsed since both samples had the same num

Accepted Answer

6 ln 2. X years ago, both had the same count N0. So N1 = N0*(1/2)^(X/3) and N2 = N0*(1/2)^(X/2). The ratio N1/N2 = (1/2)^(X/3 - X/2) = 2. Solving: (X/3 - X/2) = -1 => -X/6 = -1 => X = 6 years. Wait - let me recheck: (1/2)^(-X/6) = 2 => 2^(X/6) = 2 => X/6 = 1 => X = 6. But in terms of ln: X = 6 ln 2 only if half-life formula used in natural log form. With integer solution X = 6 years.

Question 14

The total binding energies of the nuclei He-4, Li-7, C-12, and N-14 are 28 MeV, 52 MeV, 90 MeV, and 98 MeV respectively. Which nucleus is the most stable?

Accepted Answer

C-12. Binding energy per nucleon: He-4: 28/4 = 7.0 MeV, Li-7: 52/7 = 7.43 MeV, C-12: 90/12 = 7.5 MeV, N-14: 98/14 = 7.0 MeV. C-12 has the highest binding energy per nucleon (7.5 MeV) and is therefore the most stable.

Question 15

The half-life of a radioactive substance is T0. At t = 0, the number of active nuclei is N0. Which of the following statements is correct?

Accepted Answer

The number of nuclei that have decayed in the time interval 0 to t is N0 * (1 - e^(-lambda*t)). At time t, the number of active nuclei is N(t) = N0 * e^(-lambda*t). Nuclei decayed = N0 - N(t) = N0*(1 - e^(-lambda*t)), confirming option B. For a single nucleus, probability of surviving (not decaying) = e^(-lambda*t), making option C also correct. However, option B is unambiguously correct, and option C is also correct. In standard MCQ context (single correct), both B and C are right. The question likely expects both B and C as correct (multi-correct type).

Question 16

Neon-23 undergoes beta-minus decay: 10Ne²³ -> 11Na²³ + beta⁻ + antineutrino. The atomic mass of Ne-23 is 22.9945 u and that of Na-23 is 22.9898 u. Given that 1 u = 931.5 MeV/c² and mass of electron = 0.511 MeV/c², find the maximum kinetic energy of the beta particle (in MeV), to the neares

Accepted Answer

4. For beta-minus decay using atomic masses: Q = [M(Ne-23) - M(Na-23)] * 931.5 MeV/u. The electron masses cancel because the atomic mass includes electron masses. Mass difference = 22.9945 - 22.9898 = 0.0047 u. Q = 0.0047 * 931.5 = 4.378 MeV ~ 4 MeV. The maximum kinetic energy of the beta particle equals Q (when antineutrino gets zero energy).

Question 17

A nucleus of mass number 200 breaks into two fragments with mass numbers 80 and 120. The binding energy per nucleon for the parent nucleus is 6.5 MeV, while for the two daughter fragments it is 7 MeV and 8 MeV respectively. What is the energy released (in MeV) in this nuclear fission?

Accepted Answer

200. Binding energy of parent = 200 x 6.5 = 1300 MeV. Binding energy of fragment of mass 80 = 80 x 7 = 560 MeV. Binding energy of fragment of mass 120 = 120 x 8 = 960 MeV. Energy released = (560 + 960) - 1300 = 1520 - 1300 = 220 MeV. Closest option is 200 MeV.

Question 18

Match each nuclear process in List-I with the correct nuclear equation in List-II. List-I P. Alpha decay Q. Beta-plus (positron) decay R. Nuclear fission S. Proton emission List-II 1. N-15 → N-14 +... 2. U-238 → Th-234 +... 3. Bi-185 → Pb-184 +... 4. Pu-239 → La-140 +...

Accepted Answer

P -> 2; Q -> 1; R -> 4; S -> 3. P (Alpha): U-238 -> Th-234 + He-4. This matches equation 2. R (Fission): Pu-239 splits into La-140 + smaller fragments, matches 4. Q (Beta-plus): a nucleus where Z decreases by 1, A unchanged — equation 1 could represent this if it is O-15 -> N-15 (A=15, Z drops from 8 to 7). S (Proton emission): Bi-185 -> Pb-184 + proton, Z drops from 83 to 82, A drops by 1: matches equation 3. So P->2, Q->1, R->4, S->3.

Question 19

Statement 1: Experiments show that the distance of closest approach to a gold nucleus for an alpha particle of kinetic energy 5.5 MeV is about 4 * 10^(-14) m. Statement 2: In Rutherford's scattering calculations, both the Coulombic repulsion force and the short-range nuclear force are take

Accepted Answer

Statement 1 is true; Statement 2 is false.. Statement 1 is TRUE: using energy conservation, at closest approach all kinetic energy converts to electrostatic PE: KE = k*q1*q2/r. For alpha (Z=2) and gold (Z=79): r = k*(2e)*(79e)/KE = (9e9 * 2 * 79 * (1.6e-19)²)/(5.5*1.6e-13) ~ 4e-14 m. Statement 2 is FALSE: Rutherford's classical model uses only Coulombic repulsion. Short-range nuclear forces act at distances much smaller than 4e-14 m and are NOT included in his original scattering theory.

Question 20

Two fusion reactions proceed as follows: Reaction 1: H-2 + H-2 -> H-3 + H-1 + 4.0 MeV Reaction 2: H-3 + H-2 -> He-4 + n + 17.6 MeV The net result: 3 H-2 -> He-4 + H-1 + n + 21.6 MeV What is the energy released per nucleon of the reactants?

Accepted Answer

3.6 MeV. The net reaction consumes 3 deuterium nuclei (H-2), each with 2 nucleons. Total nucleons in reactants = 3 * 2 = 6. Total energy released = 21.6 MeV. Energy per nucleon = 21.6 / 6 = 3.6 MeV.

Question 21

Beta particles emitted during radioactive decay are

Accepted Answer

electrons or positrons emitted from a nucleus. Beta rays are not electromagnetic radiation (that would be gamma rays) and are not orbital electrons. They originate from nuclear transformations: in beta-minus decay a neutron converts into a proton and emits an electron plus an antineutrino; in beta-plus decay a proton converts into a neutron and emits a positron plus a neutrino. Thus beta particles are electrons or positrons emitted by the nucleus.

Question 22

Which of the following statements about nuclear processes are ALL CORRECT? (P) In alpha decay, mass number decreases by 4 and atomic number decreases by 2. (Q) In beta-plus decay, mass number remains unchanged while atomic number decreases by 1. (R) In nuclear fission, the parent nucleus b

Accepted Answer

P, Q, R and S. P: Alpha particle is He-4 (A=4, Z=2); daughter has A reduced by 4 and Z reduced by 2. TRUE. Q: Beta-plus decay converts a proton into a neutron inside the nucleus, emitting a positron (A=0, Z=+1 for positron); total A unchanged, Z of nucleus decreases by 1. TRUE. R: Fission of a heavy nucleus (e.g., U-235) typically yields two large fragments of comparable mass (asymmetric but approximately equal). TRUE at NCERT textbook level. S: Proton emission: daughter has A-1 and Z-1. TRUE.

Question 23

Consider the nuclear fission reaction: X²⁰⁰ → A¹¹⁰ + B⁹⁰. The binding energy per nucleon for X, A, and B is 7.4 MeV, 8.2 MeV, and 8.2 MeV respectively. What is the energy released in this reaction (in MeV)?

Accepted Answer

160. The energy released in a nuclear reaction equals the increase in total binding energy (more tightly bound products release energy). BE(X) = 200 * 7.4 = 1480 MeV. BE(A) = 110 * 8.2 = 902 MeV. BE(B) = 90 * 8.2 = 738 MeV. Energy released = (902 + 738) - 1480 = 1640 - 1480 = 160 MeV.

Question 24

The binding energy per nucleon (BE/A) vs mass number (A) graph follows the standard nuclear curve peaking near A=56 (iron). Using this standard curve, which of the following statements are correct? (A) Fusion of two nuclei with mass number 30 and 45 into a nucleus of mass number 75 will re

Accepted Answer

B, C and D. Standard BE/A curve: peaks at ~8.8 MeV/nucleon around A=56. For A<56, BE/A is generally increasing. For A>56, BE/A slowly decreases. (A) A=30 has BE/A~8.5, A=45 has BE/A~8.7, A=75 has BE/A~8.7 - fusion 30+45->75: product BE/A must exceed weighted average of reactants. Since 30 has lower BE/A than 75, this actually releases energy. (B) A=80 BE/A~8.7, A=40 BE/A~8.7, Q-value is small, not 256 MeV. Statement B about exact Q-value requires reading from the provided graph - if the given graph shows specific values making Q=256 MeV, B could be correct. (C) A=150 has BE/A~8.3 MeV, A=75 has

Question 25

In an alpha-decay, the kinetic energy of the alpha particle is 48 MeV and the Q-value of the reaction is 50 MeV. The mass number of the parent nucleus is 20n. Find n. (Assume the daughter nucleus is in the ground state.)

Accepted Answer

5. In alpha decay, momentum is conserved: m_alpha * v_alpha = M_daughter * V_daughter. Kinetic energies: KE_alpha = (1/2)*m_alpha*v_alpha², KE_daughter = (1/2)*M_daughter*V_daughter². Since momenta are equal: KE_alpha/KE_daughter = M_daughter/m_alpha. Total Q = KE_alpha + KE_daughter = KE_alpha(1 + m_alpha/M_daughter). So KE_alpha = Q * M_daughter/(M_alpha + M_daughter). Plugging in: 48 = 50 * A_d/(A_d + 4), where A_d is mass number of daughter. Solving: 48(A_d + 4) = 50*A_d, 48*A_d + 192 = 50*A_d, A_d = 96. Parent mass number = 96 + 4 = 100 = 20n, so n = 5.

Question 26

The isotope boron-12 (atomic mass 12.014 u) undergoes beta-minus decay to carbon-12. Carbon-12 has a nuclear excited state (C-12*) at 4.041 MeV above its ground state. If boron-12 decays directly to this excited state C-12*, what is the maximum kinetic energy of the emitted beta particle i

Accepted Answer

9.000 MeV. Q-value = (12.014 - 12.000) * 931.5 = 13.041 MeV. Decaying to C-12* at 4.041 MeV above ground leaves KE_max = 13.041 - 4.041 = 9.000 MeV for the beta particle (maximum, when neutrino gets zero KE).

Question 27

Highly energetic electrons are bombarded on a target element containing 30 neutrons. The ratio of the radius of the target nucleus to the radius of the helium nucleus is 14^(1/3). Find the atomic number of the target nucleus.

Accepted Answer

26. From the radius ratio: (A/4)^(1/3) = 14^(1/3) gives A = 56. With 30 neutrons, Z = 56 - 30 = 26, which is iron (Fe).

Question 28

In a nuclear reactor operating at 100 MW, each fission event releases 250 MeV of energy and liberates an average of 2.5 neutrons. How many neutrons are produced per second?

Accepted Answer

6.25 * 10¹⁸. Fissions per second = Power / Energy per fission = 10⁸ / (4 * 10⁻¹¹) = 2.5 * 10¹⁸. Neutrons per second = 2.5 * 2.5 * 10¹⁸ = 6.25 * 10¹⁸.

Question 29

The nuclear binding energy per nucleon (B/A) versus mass number (A) graph is as shown. A nucleus with mass number A = 110 undergoes fission and splits into two fragments. Which of the following pairs of resulting mass numbers is energetically most favourable (i.e., releases maximum energy)

Accepted Answer

60 and 50. The B/A curve peaks around A = 56 at approximately 8.8 MeV per nucleon and decreases for both lighter and heavier nuclei. For fission of A=110 (B/A roughly 8.5 MeV), the released energy per nucleon is proportional to (B/A)_products - (B/A)_parent. The pair 60 and 50 both lie near the peak of the curve, giving the highest average B/A for the products and hence the maximum energy release.

Question 30

In a radioactive decay chain, nucleus X decays to 3 nuclei of Y plus particles and energy (decay constant 2*lambda). Each Y nucleus decays to 2 nuclei of Z plus particles and energy (decay constant lambda). Initially, only X is present with mass M (so initial number of X nuclei N0 = M/m_X)

Accepted Answer

0.06 m. At maximum, dN_Y/dt = 0: 6*lambda*N_X = lambda*N_Y, so N_Y = 6*N_X. At t=ln(10)/lambda, N_X = N0*e^(-2*ln10) = N0/100. N_Y = 6*N0/100 = 0.06*N0. Mass of Y = 0.06 * (m_Y/m_X)*M = 0.06*M if m_Y = m_X (or simply 0.06*m where m represents initial mass M and m_Y ~ m_X). Hence mass = 0.06 m.

Question 31

Carbon-11 (¹¹₆ C) decays spontaneously into Boron-11 (¹¹₅ B). Given: Atomic mass of ¹¹₆ C = 11.011434 u, Atomic mass of ¹¹₅ B = 11.009305 u, mass of electron = 0.000549 u, 1 u = 930 MeV/c². Which of the following alternatives are correct?

Accepted Answer

A beta⁺ particle is released. Since Z decreases from 6 to 5, this is beta⁺ (positron) decay. Q = [M(C-11) - M(B-11) - 2*mₑ]*c² = [11.011434 - 11.009305 - 2*0.000549]*930 MeV = [0.002129 - 0.001098]*930 = 0.001031*930 = 0.959 MeV. So both options B (beta⁺ released) and C (max KE = 0.959 MeV) are correct. In beta⁺ decay, a neutrino (not anti-neutrino) is released. So D is wrong.

Question 32

The binding energy per nucleon of a nucleus X with mass number 4 is 8 MeV. It absorbs a neutron with kinetic energy 2 MeV and transforms into nucleus Y, emitting a photon of energy 5 MeV. Find the binding energy per nucleon of Y (in MeV).

Accepted Answer

4. X (A=4): total BE = 4*8 = 32 MeV. Y = X + neutron, so A(Y) = 5. Energy conservation: BE(Y) = BE(X) + B.E. of free neutron (0) + KEₙ - E_gamma = 32 + 0 + 2 - 5 = 29 MeV. BE per nucleon of Y = 29/5 = 5.8 MeV. This doesn't match given options. Reconsidering: perhaps the question involves Q-value. Q = BE(Y) - BE(X) - 0 = KEₙ - E_gamma (net energy release in kinetic form). So BE(Y) - 32 = 2 - 5 = -3, giving BE(Y) = 29 MeV, per nucleon = 29/5. Still ~5.8. If instead BE per nucleon of Y = (32 + 2 - 5 - neutron_separation_energy)/5 or if A=4 to A=4 (no mass change), perhaps answer = 29/5 rounded or

Question 33

Consider a nucleus with mass number A = 110. Referring to a graph of binding energy per nucleon (B/A) vs mass number A, fission splits this nucleus into two fragments. Which pair of mass numbers for the resulting fragments is most energetically favourable (releases energy)?

Accepted Answer

60 and 50. The B/A vs A curve peaks around A ~ 56. Fragments with A = 60 and A = 50 are both near the peak and thus have higher B/A than the parent at A = 110, making fission energetically favourable. Pairs like 80+30, 100+10, or 90+20 have one fragment far from the peak with lower B/A.

Question 34

Carbon-11 (atomic mass 11.011434 u) undergoes decay to Boron-11 (atomic mass 11.009305 u). Given: electron mass = 0.000549 u, 1 u = 930 MeV/c². Select the correct statements about this nuclear decay.

Accepted Answer

Beta-plus particle is emitted.. C-11 (Z=6) → B-11 (Z=5): Z decreases by 1, so it is beta-plus decay, emitting a positron and a neutrino (not anti-neutrino). Q = (11.011434 - 11.009305 - 2*0.000549)*930 = (0.002129 - 0.001098)*930 = 0.001031*930 ≈ 0.959 MeV. Statements B and C are correct.

Question 35

Mx and My denote the atomic masses of the parent and daughter nuclei respectively in a radioactive decay. The Q-value for beta-minus decay is Q1 and for beta-plus decay is Q2. If me is the mass of an electron, which statement is correct?

Accepted Answer

Q1 = (Mₓ - M_y)c² and Q2 = (Mₓ - M_y - 2mₑ)c². Using atomic masses: Q1 = (Mx-My)c² because the emitted beta-minus electron is included in the Mx vs My(atomic) comparison. For beta-plus, one extra positron mass and an electron deficit in the daughter contribute, giving Q2 = (Mx-My-2me)c².

Question 36

Nucleus A has mass number 220 and binding energy per nucleon of 5.6 MeV. It splits into fragment B (mass number 104) and fragment C (mass number 116). The binding energy per nucleon in both B and C is 6.4 MeV. Find the energy Q released per fission.

Accepted Answer

176 MeV. Total BE of A = 220 * 5.6 = 1232 MeV. Total BE of B + C = 220 * 6.4 = 1408 MeV. Q = 1408 - 1232 = 176 MeV.

Question 37

The alpha-decay of nucleus X²³⁰ produces two groups of alpha-particles: group 1 with kinetic energy 6.8 MeV (daughter in ground state) and group 2 with kinetic energy 5.2 MeV (daughter in excited state). Find the energy of the gamma photon emitted during de-excitation of the daughter nucle

Accepted Answer

1.63 MeV. The excitation energy of the daughter = Q1 - Q2. Since Q = KE_alpha*(A/(A-4)) for alpha decay from nucleus of mass number A=230: Q1 = 6.8*(230/226) ≈ 6.92 MeV, Q2 = 5.2*(230/226) ≈ 5.29 MeV. E_gamma = Q1-Q2 = 1.6*(230/226) ≈ 1.63 MeV.

Question 38

Neon-23 undergoes beta-minus decay: Ne-23(10) -> Na-23(11) + beta⁻ + anti-neutrino. The atomic masses are: Ne-23 = 22.9945 u, Na-23 = 22.9898 u. Mass of electron = 0.51 MeV/c². Maximum kinetic energy of the beta⁻ particle (to nearest integer, in MeV). [Use 1 u = 931.5 MeV/c²]

Accepted Answer

4. In beta-minus decay, using atomic masses, Q = [M_parent - M_daughter] * 931.5 MeV/u. The electron masses are already included in atomic masses and cancel. Q = (22.9945 - 22.9898) * 931.5 = 0.0047 * 931.5 ≈ 4.38 MeV ≈ 4 MeV.

Question 39

The radionuclide C-11 (atomic number 6) decays by beta+ emission. Given: atomic mass of C-11 = 11.011434 u, atomic mass of B-11 = 11.009305 u, electron mass = 0.000548 u, and 1 u = 931.5 MeV/c². What is the Q-value of this decay?

Accepted Answer

0.962 MeV. In beta+ decay using atomic masses: Q = [m(C-11) - m(B-11) - 2mₑ]*c². Deltaₘ = 11.011434 - 11.009305 - 0.001096 = 0.001033 u. Q = 0.001033 * 931.5 = 0.962 MeV.

Question 40

A radioactive source contains two isotopes of phosphorus: P-32 with half-life T1 and P-33 with half-life T2 (T2 > T1). Initially 10% of all decays come from P-33. How long must one wait until 90% of the decays come from P-33?

Accepted Answer

t = 4*ln(3) / (ln(2) * (1/T1 - 1/T2)). Let lambda1 = ln2/T1, lambda2 = ln2/T2. Initially A33/A32 = 1/9. At time t: A33(t)/A32(t) = (A33₀/A32₀)*exp(-(lambda2 - lambda1)*t) = (1/9)*exp((lambda1 - lambda2)*t) = 9. So exp((lambda1-lambda2)*t) = 81, giving t = ln(81)/(lambda1-lambda2) = 4*ln3 / ((ln2/T1 - ln2/T2)) = 4*ln3/(ln2*(1/T1 - 1/T2)).

Question 41

A radioactive substance undergoes simultaneous alpha and beta decay. The mean life for alpha emission is 1620 years and for beta emission is 405 years. After how much time does only 1/4 of the original material remain?

Accepted Answer

449 years. lambda_alpha = 1/1620 yr⁻¹, lambda_beta = 1/405 yr⁻¹. lambda_total = 1/1620 + 4/1620 = 5/1620 = 1/324 yr⁻¹. Mean life tau_total = 324 yr. For N/N0 = 1/4: e^(-t/324) = 1/4 => t = 324*ln(4) = 324 * 2*ln(2) ≈ 324*1.386 ≈ 449 years.

Question 42

Tritium (³₁H), an isotope of hydrogen with a half-life of 12.3 years, is produced in the upper atmosphere and mixed with hydrogen in air. A bottle of wine found in a cave is tested and its tritium content is found to be 6.9% of that in new wine. How old is the wine?

Accepted Answer

47.5 years. Radioactive decay: N/N0 = (1/2)^(t/12.3) = 0.069. Solving: t = 12.3 * ln(1/0.069)/ln(2) = 12.3 * ln(14.49)/0.693 = 12.3 * 2.674/0.693 ~ 47.5 years.

Question 43

An unstable nucleus can decay by three different modes with individual half-lives T1, T2, and T3 where T1 >> T2 >> T3. What is the overall half-life T of the nucleus?

Accepted Answer

T ≈ T3. Total decay constant lambda = 1/T1 + 1/T2 + 1/T3 ≈ 1/T3 (since T3 is smallest, 1/T3 is largest). Overall half-life T = ln2/lambda ≈ T3.

Question 44

Substance X has a half-life of 45 years and decays into substance Y. A meteorite sample contains 2% of X and 14% of Y by quantity. Assuming Y was absent initially and all Y came from X, what is the approximate age of the meteorite?

Accepted Answer

135 years. Initial X = 2 + 14 = 16 units. Remaining X = 2 units. Fraction = 2/16 = 1/8 = (1/2)³. Number of half-lives = 3. Age = 3 * 45 = 135 years.

Question 45

A radioactive sample decays by beta emission. In the first 2 seconds, n beta-particles are emitted. In the next 2 seconds, 0.25n beta-particles are emitted. What is the half-life of the radioactive nuclei?

Accepted Answer

1 sec. Let x = e^(-2*lambda). Decays in first 2s = N₀*(1-x) = n. Decays in next 2s = N₀*x*(1-x) = 0.25n. Dividing: x = 0.25 = 1/4. So e^(-2*lambda) = 1/4 => lambda = ln2 s⁻¹. T_(1/2) = ln2/lambda = ln2/ln2 = 1 s.

Question 46

The radionuclide Carbon-11 (mass 11.011434 u) undergoes beta-plus (positron) emission to form Boron-11 (mass 11.009305 u). Given electron mass = 0.000548 u and 1 u = 931.5 MeV/c², find the Q-value of this decay.

Accepted Answer

0.962 MeV. Q = [m(C-11) - m(B-11) - 2*mₑ] * 931.5 MeV = [11.011434 - 11.009305 - 2*0.000548] * 931.5 = [0.001033] * 931.5 ≈ 0.962 MeV.

Question 47

Nuclei of a radioactive element are produced at a constant rate alpha per second. The element has decay constant lambda. Initially N₀ nuclei are present. Find the number of nuclei N at time t.

Accepted Answer

(1/lambda) * [alpha - (alpha - N₀*lambda)*e^(-lambda*t)]. The ODE dN/dt = alpha - lambda*N has solution N(t) = alpha/lambda + (N₀ - alpha/lambda)*e^(-lambda*t) = (1/lambda)*[alpha - (alpha - N₀*lambda)*e^(-lambda*t)], which matches option A.

Question 48

The electrostatic energy of Z protons uniformly distributed in a nucleus of radius R is E = 3*Z*(Z-1)*e² / (5*4*pi*epsilon₀*R). Given: mass of neutron = 1.008665 u, mass of H-1 = 1.007825 u, mass of N-15 = 15.000109 u, mass of O-15 = 15.003065 u, 1u = 931.5 MeV/c², e²/(4*pi*epsilon₀) = 1.4

Accepted Answer

3.42 fm. Delta_BE = (mₙ - m_H + M_O - M_N)*931.5 = (0.000840 + 0.002956)*931.5 = 3.54 MeV. Coulomb difference = 3*1.44*(56-42)/(5R) = 3*1.44*14/(5R) = 12.10/R. Setting equal: R = 12.10/3.54 = 3.42 fm.

Question 49

A nucleus has a radius of 7.2 * 10⁻¹⁵ m. It undergoes alpha-decay, and the daughter nucleus has a neutron-to-proton ratio of 65/41. Which of the following statements is/are correct? (A) Parent nucleus is 84Po. (B) Daughter nucleus is 82Pb. (C) Mass number of daughter nucleus is 216. (D) Ma

Accepted Answer

Parent nucleus is 84Po.. Parent: R = R0*A^(1/3) gives A = 216. After alpha decay: daughter A = 212, Z = 82 (Pb-212), parent Z = 84 (Po-216). Options A, B, D are correct; C is incorrect (daughter mass number is 212, not 216).

Question 50

A radioactive sample has a half-life of 40 seconds. Its activity 80 seconds after the start is measured to be 6.932 * 10¹⁸ dps. The total energy released during this 80-second interval is 6 * 10⁸ J (given ln2 = 0.6932). Which of the following are correct?

Accepted Answer

The initial number of atoms in the sample is 1.6 * 10²¹. lambda = 0.6932/40 = 0.01733 s⁻¹. Activity after 2 half-lives: A(80) = lambda*N0/4. So N0 = 4*A(80)/lambda = 4 * 6.932e18 / 0.01733 = 1.6e21. Total decays = 3N0/4 = 1.2e21. Energy per decay = 6e8/1.2e21 = 5e-13 J. Options B and C are correct.

JEE Advanced Physics: Nuclei questions with solutions

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