DC Pandey Solutions Modern Physics II - Physics Class 12 - NEET PDF Download

Introductory Exercise 31.1

Ques 1: Activity of a radioactive substance decreases from 8000 8q to 1000 Bq in 9 days. What is the half life and average life of the radioactive substance?
Sol: 

Initial activity = 8000 Bq, final activity = 1000 Bq in t = 9 days.
Activity ratio = 8000/1000 = 8 = 2³.
Thus 3 half-lives occur in 9 days, so t_1/2 = 9/3 = 3 days.
Mean life (average life) τ = t_av = 1.44 × t_1/2 = 1.44 × 3 = 4.32 days.
Therefore, half-life = 3 days and average life = 4.32 days.
Ques 2: A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity  (t = 0) of 40 μCi. Calculate the number of nuclei that decay in the time interval between t₁ = 10.0 h and t₂ =12.0 h.
Sol: R₀ = λN₀.
First find λ: λ = ln 2 / T_1/2 = 0.693147 / (64.8 × 3600 s) = 2.971 × 10^-6 s^-1.
Convert initial activity to Bq: 40 μCi = 40 × 10^-6 Ci = 40 × 10^-6 × 3.7 × 10¹⁰ s^-1 = 1.48 × 10⁶ Bq.
Number of nuclei at t = 0: N₀ = R₀/λ = (1.48 × 10⁶)/(2.971 × 10^-6) ≈ 4.98 × 10¹¹.
Number decayed between t₁ and t₂: ΔN = N(t₁) - N(t₂) = N₀(e^-λt1 - e^-λt2).
Compute λt₁ = 2.971 × 10^-6 × 36 000 = 0.1070, e^-λt1 ≈ 0.8986.
Compute λt₂ = 2.971 × 10^-6 × 43 200 = 0.1283, e^-λt2 ≈ 0.8794.
ΔN ≈ 4.98 × 10¹¹ × (0.8986 - 0.8794) ≈ 9.6 × 10⁹ nuclei.
Thus about 9.6 × 10⁹ nuclei decay between 10.0 h and 12.0 h.
Ques 3: A freshly prepared sample of a certain radioactive isotope has an activity of 10 mCi. After 4.0 h its activity is 8.00 mCi.
(a) Find the decay constant and half life
(b) How many atoms of the isotope were contained in the freshly prepared sample?
(c) What is the sample's activity 30.0 h after it is prepared?
Sol: (a) Use R = R₀ e^-λt.
8.00 = 10.0 e^-λ×4 ⇒ e^-4λ = 0.8 ⇒ -4λ = ln 0.8 = -0.223143 ⇒ λ = 0.05579 h^-1.
Half-life t_1/2 = ln 2 / λ = 0.693147 / 0.05579 ≈ 12.43 h.
(b) Convert R₀ to Bq: 10 mCi = 10 × 10^-3 Ci = 0.01 Ci = 0.01 × 3.7 × 10¹⁰ s^-1 = 3.7 × 10⁸ Bq.
Convert λ to s^-1: λ = 0.05579 / 3600 = 1.5497 × 10^-5 s^-1.
Number of atoms N₀ = R₀/λ = (3.7 × 10⁸)/(1.5497 × 10^-5) ≈ 2.39 × 10¹³ atoms.
(c) Activity after t = 30.0 h: R = R₀ e^-λt = 10 mCi × e^-0.05579×30.
Exponent = -1.6737, e^-1.6737 ≈ 0.1876 ⇒ R ≈ 1.88 mCi.
Thus (a) λ ≈ 0.05579 h^-1, t_1/2 ≈ 12.43 h; (b) N₀ ≈ 2.39 × 10¹³; (c) R(30 h) ≈ 1.88 mCi.
Ques 4: A radioactive substance contains 10¹⁵ atoms and has an activity of 6.0 × 10¹¹ Bq. What is its half-life?
Sol: R₀ = λN₀ ⇒ λ = R₀/N₀ = (6.0 × 10¹¹)/(10¹⁵) = 6.0 × 10^-4 s^-1.
Half-life t_1/2 = ln 2 / λ = 0.693147 / (6.0 × 10^-4) ≈ 1.155 × 10³ s ≈ 19.25 min.
Hence t_1/2 ≈ 1155 s ≈ 19.25 minutes.


Ques 5: Two radioactive elements X and Y have half-life periods of 50 minutes and 100 minutes respectively. Initially both of them contain equal number of atoms. Find the ratio of atoms left N_X/N_Y after 200 minutes.
Sol: For X: number of half-lives n_X = 200/50 = 4 ⇒ fraction left = (1/2)⁴ = 1/16.
For Y: n_Y = 200/100 = 2 ⇒ fraction left = (1/2)² = 1/4.
Initially numbers equal, so N_X/N_Y = (1/16)/(1/4) = 1/4.
Therefore the ratio N_X/N_Y = 1 : 4 (or 0.25). 

Introductory Exercise 31.2

Ques 1: (a) How much mass is lost per day by a nuclear reactor operated at a 10⁹ watt power level?
(b) If each fission releases 200 MeV, how many fissions occur per second to yield this power level?
Sol:  (a) Energy produced per day P × t = 10⁹ W × 86 400 s = 8.64 × 10¹³ J.
Mass equivalent m = E / c² ≈ (8.64 × 10¹³)/(8.9876 × 10¹⁶) ≈ 9.61 × 10^-4 kg = 0.961 g.
So about 9.6 × 10^-4 kg (≈ 0.96 g) of mass is converted per day.
(b) Energy per fission = 200 MeV = 200 × 1.602 × 10^-13 J = 3.204 × 10^-11 J.
Fissions per second = Power / energy per fission = (10⁹)/(3.204 × 10^-11) ≈ 3.12 × 10¹⁹ fissions s^-1.
Therefore about 3.12 × 10¹⁹ fissions per second are required.

(b) Number of fissions required per second

= 3.125 × 10¹⁹
Ques 2: Find energy released in the alpha decay

Given

Sol: Mass defect Δm = ∑m_initial - ∑m_final = (238.050784) - (234.043593 + 4.002602) = 4.589 × 10^-3 u.
Energy released Q = Δm × 931.48 MeV = 4.589 × 10^-3 × 931.48 ≈ 4.27 MeV.
Thus the alpha-decay releases ≈ 4.27 MeV of energy.

Exercises
For JEE Main

Subjective Questions
Radioactivity
Ques 1: The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. Five minutes later the rate becomes 2700 per minute. Calculate
(a) decay constant and
(b) half-life of the sample
Sol: (a) Use R = R₀ e^-λt.
Given R₀ = 4750 min^-1, R = 2700 min^-1, t = 5 min.
λ = (1/t) ln(R₀/R) = (1/5) ln(4750/2700) = (1/5) ln(1.75926) = 0.11287 min^-1.
(b) t_1/2 = ln 2 / λ = 0.693147 / 0.11287 ≈ 6.14 min.
Therefore λ ≈ 0.1129 min^-1 and t_1/2 ≈ 6.14 minutes.

Ques 2: A radioactive sample contains 1.00 × 10¹⁵ atoms and has an activity of 6.00 × 10¹¹ Bq. What is its half-life?
Sol: Activity R = λN ⇒ λ = R/N = (6.00 × 10¹¹)/(1.00 × 10¹⁵) = 6.00 × 10^-4 s^-1.
Half-life t_1/2 = ln 2 / λ = 0.693147 / (6.00 × 10^-4) ≈ 1.155 × 10³ s = 19.25 min.
 = 1155 s = 19.25 min
Ques 3: Obtain the amount of ⁶⁰Co necessary to provide a radioactive source of 8.0 Ci strength. The half-life of ⁶⁰Co is 5.3 years?
Sol: Activity R = λN.
T_1/2 = 5.3 y = 5.3 × 365 × 24 × 3600 = 1.671408 × 10⁸ s.
λ = ln 2 / T_1/2 = 0.693147 / 1.671408 × 10⁸ ≈ 4.149 × 10^-9 s^-1.
Activity R = 8.0 Ci = 8 × 3.7 × 10¹⁰ s^-1 = 2.96 × 10¹¹ Bq.
Number of atoms required N = R / λ = (2.96 × 10¹¹)/(4.149 × 10^-9) ≈ 7.14 × 10¹⁹ atoms.
Mass of 60Co: atomic mass ≈ 60 g mol^-1, so mass m = N × (60 g mol^-1)/(6.022 × 10²³ atoms mol^-1)
= 7.14 × 10¹⁹ × (60/6.022 × 10²³) g ≈ 7.11 × 10^-3 g = 7.11 mg.
Therefore about 7.14 × 10¹⁹ atoms or ≈ 7.1 mg of 60Co is needed.

= 7.14 × 10¹⁹

Ques 4: The half-life of

 against alpha decay is 4.5 × 10⁹ year. How much disintegration per second occurs in 1 g of

?
Sol: 
Take T_1/2 = 4.5 × 10⁹ y = 4.5 × 10⁹ × 3.1536 × 10⁷ s = 1.41912 × 10¹⁷ s.
λ = ln 2 / T_1/2 = 0.693147 / 1.41912 × 10¹⁷ = 4.884 × 10^-18 s^-1.
Number of atoms in 1 g (assume mass number ≈ 238): N = (1 g)/(238 g mol^-1) × 6.022 × 10²³ ≈ 2.529 × 10²¹ atoms.
Activity R = λN = (4.884 × 10^-18) × (2.529 × 10²¹) ≈ 1.235 × 10⁴ s^-1.
Thus about 1.23 × 10⁴ disintegrations per second occur in 1 g.

= 1.23 × 10⁴ dps
5. 1/λ = 10days
∴ λ = 0.1 day^-1
Probability of decay

= 1 - e^-λt = 1 - e-^0.1×5 = 0.39
Ques 5: In an ore containing Uranium, the ratio of ²³⁸U to ²⁰⁶Pb nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of ²³⁸U. Take the half-life of ²³⁸U to be 4.5 × 10⁹ years.
Sol: If current ratio N_U:N_Pb = 3:1, then initial uranium atoms N₀ = N_U + N_Pb = 3 + 1 = 4 (in relative units).
Fraction of uranium remaining = N/N₀ = 3/4 = e^-λt.
So t = (1/λ) ln(N₀/N) = (1/λ) ln(4/3).
λ = ln 2 / T_1/2 = 0.693147 / (4.5 × 10⁹ y) = 1.5403 × 10^-10 y^-1.
t = (ln(4/3))/λ = 0.287682 / (1.5403 × 10^-10) ≈ 1.87 × 10⁹ years.
Therefore the age of the ore ≈ 1.87 × 10⁹ years.

N₀ = 3 + 1 = 4 N =3

From Eqs. (i) and (ii), we get t = 1.88 × 10⁹ yr
Ques 6: The half-lives of radioisotopes P³² and P³³ are 14 days and 25 days respectively. These radioisotopes are mixed in the ratio of 4 :1 of their atoms. If the initial activity of the mixed sample is 3.0 m Ci, find the activity of the mixed isotopes after 60 year.
Sol:

Interpreting the intended time as 60 days (typical context for these half-lives):
Let initial numbers be N1₀ : N2₀ = 4 : 1.
Decay constants: λ1 = ln 2 / 14 = 0.04951 day^-1, λ2 = ln 2 / 25 = 0.02773 day^-1.
Initial activities A1₀ ∝ λ1 N1₀, A2₀ ∝ λ2 N2₀.
Let N2₀ = k ⇒ N1₀ = 4k. Then A1₀/A2₀ = (λ1×4k)/(λ2×k) = 4(λ1/λ2) = 4 × (0.04951/0.02773) ≈ 7.14.
Total initial activity 3.0 mCi ⇒ A1₀ = (7.14/8.14) × 3.0 ≈ 2.63 mCi, A2₀ ≈ 0.37 mCi.
After t = 60 days: A1 = A1₀ e^{-λ1 t}, A2 = A2₀ e^{-λ2 t}.
λ1 t = 0.04951 × 60 = 2.9706 ⇒ e^-2.9706 = 0.0514 ⇒ A1 ≈ 2.63 × 0.0514 = 0.135 mCi.
λ2 t = 0.02773 × 60 = 1.6638 ⇒ e^-1.6638 = 0.1895 ⇒ A2 ≈ 0.37 × 0.1895 = 0.070 mCi.
Total activity after 60 days ≈ 0.205 mCi.
Thus the mixed sample activity after 60 days ≈ 0.205 mCi (≈ 205 μCi).
Ques 7: Consider two decay reactions.

 Pb+ 10 protons + 20 neutrons

Are both the reactions possible?
Sol: (a) Check conservation: initial Z + added protons = 82 + 10 = 92; initial A + added nucleons = 206 + 10 + 20 = 236. Both atomic number and mass number balance, so the reaction is possible as written (conservation of nucleon number and charge holds).
(b) For the second reaction the numbers give 82 + 16 - 6 = 92 and 206 + 32 = 238 in the final accounting shown. As written, charge and nucleon balances do not match without specifying the emitted particle(s). This reaction would require the emission or absorption of additional particles (for example β⁺ / electron capture and an associated neutrino/antineutrino) to conserve charge and lepton number. Therefore, as stated, the second reaction is not possible unless the required additional particle(s) (such as a positron and neutrino or an electron and antineutrino) are explicitly included.
Ques 8: Obtain the binding energy of a nitrogen nucleus from the following data :
m_H = 1.00783 u, m_N = 1.00867 u, m(¹⁴₇ N) = 14.00307 u
Give your answer in units of MeV. [Remember 1 u = 931.5 Me V/c²]
Sol: Mass of 7 protons = 7 × 1.00783 = 7.05481 u.
Mass of 7 neutrons = 7 × 1.00867 = 7.06069 u.
Sum of separate nucleon masses = 7.05481 + 7.06069 = 14.11550 u.
Mass defect Δm = 14.11550 - 14.00307 = 0.11243 u.
Binding energy = Δm × 931.5 MeV = 0.11243 × 931.5 ≈ 104.7 MeV.
Therefore the binding energy of ¹⁴N ≈ 104.7 MeV.