Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.
Try each problem first, then open the solution to check your working.
Find the radius of a nucleus of mass number 64.
Given: A = 64; R₀ = 1.2 fm
Use: R = R₀ A^(1/3)
R = 1.2 × 4 = 4.8 fm
Answer: 4.8 fm
A nucleus has a radius of 3.6 fm. Find its mass number.
Given: R = 3.6 fm
Use: A = (R/R₀)³
A = (3.6/1.2)³ = 27
Answer: 27
Estimate the density of nuclear matter.
Given: R₀ = 1.2 fm; 1 u = 1.66 × 10⁻²⁷ kg
Use: ρ = A u / (4/3 π R₀³ A)
ρ = 1.66 × 10⁻²⁷ / [4/3 π (1.2 × 10⁻¹⁵)³] = 2.3 × 10¹⁷ kg/m³
Answer: 2.3 × 10¹⁷ kg/m³
Find the ratio of the radii of the nuclei of ¹⁹⁷Au and ¹⁰⁷Ag.
Given: A₁ = 197; A₂ = 107
Use: R ∝ A^(1/3)
R₁/R₂ = (197/107)^(1/3) = 1.23
Answer: 1.23
Find the binding energy of ¹⁴₇N (atomic mass 14.00307 u). Use m_H = 1.007825 u, mₙ = 1.008665 u.
Given: Z = 7; N = 7
Use: ΔM = 7m_H + 7mₙ − M
ΔM = 7.054775 + 7.060655 − 14.00307 = 0.11236 u
E_b = 0.11236 × 931.5 = 104.7 MeV
Per nucleon: 104.7/14 = 7.48 MeV
Answer: 104.7 MeV; 7.48 MeV per nucleon
Find the binding energy and the binding energy per nucleon of ⁵⁶₂₆Fe (55.934939 u) and ²⁰⁹₈₃Bi (208.980388 u).
Given: As given
Use: E_b = [Zm_H + (A − Z)mₙ − M] × 931.5
Fe: ΔM = 0.52846 u, E_b = 492.3 MeV, 8.79 MeV per nucleon
Bi: ΔM = 1.76088 u, E_b = 1640.3 MeV, 7.85 MeV per nucleon
Answer: Fe: 492.3 MeV (8.79/nucleon); Bi: 1640.3 MeV (7.85/nucleon)
Find the energy released in joule if 1 g of matter is completely converted to energy.
Given: m = 10⁻³ kg; c = 3 × 10⁸ m/s
Use: E = mc²
E = 10⁻³ × 9 × 10¹⁶ = 9 × 10¹³ J
Answer: 9 × 10¹³ J
In a fission event 0.2 u of mass is converted into energy. Find the energy released.
Given: Δm = 0.2 u
Use: E = Δm × 931.5 MeV
E = 186 MeV
Answer: 186 MeV
Find the Q-value of (i) ¹₁H + ³₁H → ²₁H + ²₁H and (ii) ¹²₆C + ¹²₆C → ²⁰₁₀Ne + ⁴₂He. Masses (u): ²₁H 2.014102; ³₁H 3.016049; ¹H 1.007825; ¹²C 12.000000; ²⁰Ne 19.992439; ⁴He 4.002603.
Given: As given
Use: Q = (Σm_i − Σm_f) × 931.5
(i) Δm = 1.007825 + 3.016049 − 4.028204 = −0.00433 u, Q = −4.03 MeV: endothermic
(ii) Δm = 24 − 19.992439 − 4.002603 = 0.004958 u, Q = +4.62 MeV: exothermic
Answer: (i) −4.03 MeV; (ii) +4.62 MeV
Is the fission of ⁵⁶₂₆Fe into two equal ²⁸₁₃Al fragments energetically possible? m(Fe) = 55.93494 u, m(Al) = 27.98191 u.
Given: As given
Use: Q = [m(Fe) − 2m(Al)] × 931.5
Δm = 55.93494 − 55.96382 = −0.02888 u
Q = −26.9 MeV < 0
Answer: Not possible, since Q is negative
The average energy released per fission of ²³⁹Pu is 180 MeV. Find the energy released if all atoms in 1 kg undergo fission.
Given: m = 1 kg; E = 180 MeV
Use: E = N × 180 MeV
N = 1000/239 × 6.023 × 10²³ = 2.52 × 10²⁴
E = 2.52 × 10²⁴ × 180 = 4.5 × 10²⁶ MeV
Answer: 4.5 × 10²⁶ MeV
Find the height of the potential barrier for a head-on collision of two deuterons (hard spheres of radius 2.0 fm).
Given: r = 2.0 fm each, so d = 4.0 fm
Use: U = e²/(4πε₀ d)
U = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² / 4 × 10⁻¹⁵ = 5.76 × 10⁻¹⁴ J
= 0.36 MeV
Answer: 0.36 MeV
Write the daughter nucleus after α-decay of ²³⁸₉₂U.
Given: ²³⁸₉₂U → X + ⁴₂He
Use: A decreases by 4, Z by 2
A = 234, Z = 90, which is thorium
Answer: ²³⁴₉₀Th
A nucleus ¹⁴₆C undergoes β⁻ decay. Find the daughter.
Given: β⁻: n → p + e⁻ + antineutrino
Use: Z increases by 1, A is the same
Z = 7, A = 14: nitrogen
Answer: ¹⁴₇N
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(Data: mH = 1.007825 u, mₙ = 1.008665 u, 1 u = 931.5 MeV/c²; the given mass is the atomic mass.)
The nitrogen atom has 7 electrons and 7 protons, and 7 neutrons. Mass of constituents = 7 × 1.007825 + 7 × 1.008665 = 14.115430 u.
Mass defect ΔM = 14.115430 − 14.00307 = 0.11236 u.
E_b = 0.11236 × 931.5 = 104.7 MeV (7.48 MeV per nucleon).
Fe: ΔM = 26 × 1.007825 + 30 × 1.008665 − 55.934939 = 26.203450 + 30.259950 − 55.934939 = 0.528461 u. E_b = 0.528461 × 931.5 = 492.3 MeV; per nucleon 492.3/56 = 8.79 MeV.
Bi: ΔM = 83 × 1.007825 + 126 × 1.008665 − 208.980388 = 83.649475 + 127.091790 − 208.980388 = 1.760877 u. E_b = 1.760877 × 931.5 = 1640.3 MeV; per nucleon 1640.3/209 = 7.85 MeV.
(Fe has the greater binding energy per nucleon, so it is the more tightly bound.)
Mass defect per atom = 29 × 1.007825 + 34 × 1.008665 − 62.92960 = 29.226925 + 34.294610 − 62.92960 = 0.591935 u.
Binding energy per atom = 0.591935 × 931.5 = 551.4 MeV.
Number of atoms = (3.0 / 62.9296) × 6.023 × 10²³ = 2.87 × 10²².
Total energy = 2.87 × 10²² × 551.4 = 1.58 × 10²⁵ MeV = 1.58 × 10²⁵ × 1.6 × 10⁻¹³ J = 2.5 × 10¹² J.
R ∝ A^(1/3), so R_Au/R_Ag = (197/107)^(1/3) = (1.841)^(1/3) = 1.23.
(i) Using m(¹H) = 1.007825 u: Q = [1.007825 + 3.016049 − 2 × 2.014102] × 931.5 = (−0.004330) × 931.5 = −4.03 MeV. Q is negative: the reaction is endothermic.
(ii) Using m(⁴He) = 4.002603 u: Q = [24.000000 − 19.992439 − 4.002603] × 931.5 = 0.004958 × 931.5 = +4.62 MeV. Q is positive: the reaction is exothermic.
Q = [m(Fe) − 2 m(Al)] c² = [55.93494 − 2 × 27.98191] × 931.5 = (55.93494 − 55.96382) × 931.5 = −0.02888 × 931.5 = −26.9 MeV.
Q is negative, so energy would have to be supplied: the fission of ⁵⁶Fe is not energetically possible. (Fe lies at the peak of the binding energy curve, so splitting it lowers the binding energy per nucleon.)
Number of atoms in 1 kg = (1000 g / 239 g) × 6.023 × 10²³ = 2.52 × 10²⁴.
Energy = 2.52 × 10²⁴ × 180 MeV = 4.5 × 10²⁶ MeV.
Number of deuterium atoms in 2.0 kg = (2000 / 2.014) × 6.023 × 10²³ ≈ 5.98 × 10²⁶. (Taking the molar mass as 2.014 g.)
Each reaction uses 2 nuclei and gives 3.27 MeV, so the number of reactions = 2.99 × 10²⁶, and the energy = 2.99 × 10²⁶ × 3.27 MeV = 9.78 × 10²⁶ MeV = 1.56 × 10¹⁴ J.
Time = E / P = 1.56 × 10¹⁴ / 100 = 1.56 × 10¹² s = 1.56 × 10¹² / 3.154 × 10⁷ = 5.0 × 10⁴ years.
When the two deuterons just touch, their centres are 2 × 2.0 = 4.0 fm apart.
Barrier height = e² / (4πε₀ d) = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² / (4.0 × 10⁻¹⁵) = 5.76 × 10⁻¹⁴ J = 5.76 × 10⁻¹⁴ / 1.6 × 10⁻¹³ MeV = 0.36 MeV (360 keV).
Mass of the nucleus ≈ A u (A atomic mass units, 1 u = 1.66 × 10⁻²⁷ kg). Volume = (4/3)πR³ = (4/3)πR₀³A.
Density ρ = mass / volume = A u / [(4/3)πR₀³ A] = 3u / (4πR₀³), in which A cancels. So ρ is independent of A.
Numerically, with R₀ = 1.2 × 10⁻¹⁵ m: ρ = 3 × 1.66 × 10⁻²⁷ / [4π (1.2 × 10⁻¹⁵)³] = 2.3 × 10¹⁷ kg/m³.