StudentAidUnit 12: Nuclear and Particle Physics — Numericals
11th Class Physics · Unit 12: Nuclear and Particle Physics
12.1.Uranium-238 is an alpha emitter. In the process, it is transmutated into a daughter nucleus. What is the mass number A and charge number Z of the daughter nucleus? What is its chemical symbol?
Given
Parent nucleus
{}23892U
Formula
Alpha decay process
During α emission, atomic number decreases by 2 and atomic mass decreases by 4
Mass number of daughter
A = 238 - 4 = 234
Charge number of daughter
Z = 92 - 2 = 90
Result
Nuclear equation
{}23892U → {}23490X + {}42He
Product nucleus
Thorium-234 (Th)
12.2.Polonium-218 (Pc) is a beta minus emitter. What will be the mass number A and charge number Z of the daughter nucleus?
Given
Parent nucleus
{}21884Po
Formula
Beta negative emission
During beta negative emission, atomic number (Z) increases by 1 and mass number (A) remains same
Nuclear equation
{}21884Po → {}218X + {}0-1e
Charge number of daughter
Z = 84 + 1 = 85
Mass number of daughter
A = 218 (remains same)
Result
Product nucleus
Astatine-218 (At) with Z=85
12.3.Nitrogen-14 (N) bombarded by alpha particle result in Oxygen-17 (O). What is the product particle in this nuclear reaction? Write the nuclear reaction equation.
Given
Reactants
{}14N + {}4He → {}17O + ?
Formula
Conservation of mass number
A_{initial = A_{final
Mass number calculation
14 + 4 = 17 + x ⇒ x = 1
Conservation of charge number
Z_{initial = Z_{final
Charge number calculation
7 + 2 = 8 + y ⇒ y = 1
Result
Product particle
The particle with mass number 1 and charge number 1 is a proton ({}1H)
Complete nuclear equation
{}14N + {}4He → {}17O + {}1H
12.4.Show that nucleon number N and charge number Z are conserved in the numerical question 12.3.
Given
Nuclear reaction from 12.3
{}14N + {}4He → {}17O + {}1H
Nucleon numbers (mass numbers) on left side
14 (N) + 4 (He) = 18
Nucleon numbers on right side
17 (O) + 1 (H) = 18
Result
Nucleon conservation
Nucleon number is conserved: 18 = 18
Charge numbers on left side
7 (N) + 2 (He) = 9
Charge numbers on right side
8 (O) + 1 (H) = 9
Charge conservation
Charge number is conserved: 9 = 9
12.5.Determine the rest-mass energy of electron in eV. Its rest-mass is 0.000555u?
Given
Rest mass of electron
m0 = 0.000555 u = 9.1 × 10-31 kg
Formula
Rest mass energy formula
E = mc2
Working
Substitute values
E = (9.1 × 10-31) kg × (3 × 108)2 m2s-2
Calculate
E = (9.1 × 10-31) × (9 × 1016)
Result in joules
E = 8.29 × 10-14 J
Convert to eV (1 eV = 1.6 × 10⁻¹⁹ J)
E = frac{8.29 × 10-14{1.6 × 10-19 eV
Calculate
E = 0.511 × 106 eV
Result
Final answer
E = 0.511 MeV
12.6.What is Q value of a nuclear reaction? Calculate it for the reaction taking place in Rutherford's experiment on artificial disintegration of nitrogen by bombardment with alpha particles. Relative masses are: N-14 = 14.007515u, He-4 = 4.003837u, O-17 = 17.004533u, H-1 = 1.008142u
Result
Nuclear reaction equation
{}147N + {}42He → {}178O + {}11H
Formula
Q value formula
Q = [m(N) + m(He) - m(O) - m(H)] × c2
Given
Atomic masses (a.m.u.)
m(N) = 14.003074 u, quad m(He) = 4.002603 u, quad m(O) = 16.999132 u, quad m(H) = 1.007825 u
Calculate mass sum of reactants
14.003074 + 4.002603 = 18.005677 u
Calculate mass sum of products
16.999132 + 1.007825 = 18.006957 u
Calculate mass difference
Δm = 18.005677 - 18.006957 = -0.00128 u
Formula
Conversion factor
1 a.m.u. = 931.5 MeV
Convert to energy
Q = -0.00128 × 931.5 MeV
Calculate
Q ≈ -1.19232 MeV
Result
Final answer
Q ≈ -1.2 MeV (endothermic reaction)