Nuclear Chemistry Previous Year Questions with Solutions

1.0Introduction(Nuclear Chemistry)

Nuclear Chemistry is the study of nuclear reactions, radioactivity, and the behaviour of radioactive elements. It explores processes such as nuclear decay, fission, fusion, and the forces that hold the nucleus together.

Nuclear Chemistry is a high-scoring and concept-based topic in Physical Chemistry. It covers fundamental principles such as radioactivity, nuclear reactions, half-life, and decay laws—all of which are frequently asked in competitive exams like JEE. While Nuclear Chemistry is a relatively small chapter, 1–2 questions are typically asked each year in JEE Main. In JEE Advanced, it may be integrated into mixed-concept or assertion-reason questions.

2.0Key Concepts to Remember

Nuclear Stability & Mass Defect: Some nuclei are stable, while others undergo radioactive decay. The mass defect is the difference between the mass of a nucleus and its nucleons, converted into binding energy that holds the nucleus together. Binding energy (B.E. = Δm × c²) indicates nuclear stability—higher B.E. means greater stability.

Nuclear Forces: Nuclear forces are strong, short-range forces binding protons and neutrons via meson exchange. They are much stronger than electrostatic forces.

Radioactivity & Types: Unstable nuclei emit α, β, or γ radiation:

• Alpha (α): He nucleus, low penetration, high ionization, reduces mass by 4 and atomic number by 2.
• Beta (β): Electron or positron emission; atomic number changes by ±1.
• Gamma (γ): High-energy photons; no change in mass or atomic number.

Decay Kinetics: Radioactive decay follows first-order kinetics: $N=N_0{e}^{-kt}$

• Half-life (t₁/₂): Time for half the atoms to decay.
• Activity (A): Disintegrations per second (A = kN).
• Average life: t = 1/k.

Decay Series: Radioactive series (Thorium, Uranium, Actinium, Neptunium) transform unstable elements into stable lead or bismuth isotopes via α and β decay.

Nuclear Reactions:

• Fission: Splitting heavy nuclei (e.g., U-235) releases ~200 MeV.
• Fusion: Combining light nuclei (e.g., H isotopes) at high temperatures releases enormous energy.

Applications:

• Carbon & Uranium Dating: Used in archaeology and geology.
• Medical: Cancer treatment using radioisotopes.

3.0JEE Main Past Year Questions with Solutions on Nuclear Chemistry

Q.1 The half-life of radioisotopic bromine-82 is 36 hours. The fraction which remains after one day is ___________  ×10^–2.

(Given antilog 0.2006 = 1.587)

Ans.   (63)

Solution.  

Half life of bromine – 82 = 36 hours

$t_{1/2}=\frac{0.693}{K}$

$K=\frac{0.693}{36}=0.01925{\text{hr}}^{-1}$

1^st order reaction kinetic equation

$t=\frac{2.303}{K}\log \frac{a}{a-x}$

$\log \frac{a}{a-x}=\frac{t\times K}{2.303}(t=1\text{day}=24\text{hr})$

$\log \frac{a}{a-x}=\frac{24\text{hr}\times 0.01925{\text{hr}}^{-1}}{2.303}$

$\log \frac{a}{a-x}=0.2006$

$\frac{a}{a-x}=\text{antilog}(0.2006)$

$\frac{a}{a-x}=1.587$

If a = 1

$\therefore 1-x=0.6301=\text{Fraction remain after one day}$

Q2 . The rate of first order reaction is 0.04 mol L^–1 s^–1 at 10 minutes and 0.03 mol L^-1 s^-1 at 20 minutes after initiation. Half life of the reaction is _______ minutes. (Given log2=0.3010, log3=0.4771)

Ans. (24)

Solution

$0.04=k[A{]}_0{e}^{–k\times 10\times 60}$          …(1)

$0.03=k[A{]}_0{e}^{–k\times 20\times 60}$             …(2)

$(1)/(2)$

$\frac{4}{3}=e^{600k(2-1)}$

$\frac{4}{3}=e^{600k}$

$ln\frac{4}{3}=600k$

$ln\frac{4}{3}=600\times \frac{ln2}{t_{\frac{1}{2}}}$

$t_{\frac{1}{2}}=600\frac{ln2}{ln\frac{4}{3}}sec$

$t_{\frac{1}{2}}=600\times \frac{log2}{log4-log3}sec$

$t_{\frac{1}{2}}=10\times \frac{\mathrm{0.3.1.}}{0.6020-0.477}min$

$t_{\frac{1}{2}}=24.08min$

Ans. 24

Q. 3  At 30°C, the half-life for the decomposition of AB₂ is 200 s and is independent of the initial concentration of AB₂. The time required for 80% of the AB₂ to decompose is

 (Given: log 2 = 0.30; log 3 = 0.48)

 (1)  200 s                  (2) 323 s                      (3) 467 s              (4) 532 s 

 Ans. (3)

Sol.   T_1/2 = 200 s and 1^st order reaction

$K=\frac{2.3030log2}{200}=\frac{2.303}{t}log\frac{A_0}{0.2A_0}$

$\frac{log2}{200}=\frac{1}{t}log5$

$t=\frac{7}{3}\times 200=466.67s$

$467s$

Q.4 During the nuclear explosion, one of the products is 90Sr with a half-life of 6.93 years. If 1 g of 90Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically______

Ans. (23 to 23.03)

Sol. All nuclear decays follow first-order kinetics.

$t=\frac{1}{k}ln\frac{[A_o]}{[A]}$

$\frac{(t_{\frac{1}{2}})}{0.693}=2.303lo{g}_{10}10=10\times 2.303\times 1$

$=23.03yrs$

Q. 5 Half-life of a zero-order reaction A → product is 1 hour, when the initial concentration of the reactant is 2.0 mol L⁻¹. The time required to decrease the concentration of A from 0.50 mol L⁻¹ to 0.25 mol L⁻¹ is:

Ans: 15 minutes

Solution:

For a zero-order reaction, the rate law is:

$[A{]}_t=[A{]}_0-kt$

Also, the half-life (t₁/₂) for zero order is:

$t_{1/2}=\frac{[A{]}_0}{2k}$

Given:

• $t1/2=1t_{1/2}=1t1/2=1hour$
• $[A{]}_0=2.0{\text{mol L}}^{-1}$

Step 1: Find the rate constant k

$1=\frac{2.0}{2k}\Rightarrow k=1.0{\text{mol L}}^{-1}{\text{hr}}^{-1}$

Step 2: Use the integrated zero-order rate law:

$[A{]}_t=[A{]}_0-kt$

Let:

• $[A{]}_0=0.50{\text{mol L}}^{-1}$
• $[A{]}_t=0.25{\text{mol L}}^{-1}$

0.25=0.50−(1.0)t ⇒ t=0.25 hr = 15 minutes

Ans: 15 minutes

Q.6 The half-life period of a first-order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2=0.301) :

A. 230.3 minutes

B. 23.03 minutes

C. 46.06 minutes

D. 460.6 minutes

Ans: (C) 46.06 minutes

Given:

• Half-life of a first-order reaction: $t_{1/2}=6.93\text{minutes}$
• We need the time required to complete 99% of the reaction, i.e., when [A] = 1% of [A]₀

For first-order reactions, the integrated rate law is:

$\ln \left(\frac{[A{]}_0}{[A]}\right)=kt$

When 99% is completed:

$\frac{[A{]}_0}{[A]}=\frac{100}{1}=100\Rightarrow \ln (100)=kt$

We know:

$t_{1/2}=\frac{0.693}{k}\Rightarrow k=\frac{0.693}{6.93}=0.1{\text{min}}^{-1}$

Now,

$ln(100)=\ln (10^2)=2\ln (10)=2\times 2.303=4.606$

Now plug in:

$4.606=0.1\times t\Rightarrow t=\frac{4.606}{0.1}=46.06\text{minutes}$

Answer = 46.06 minutes

Q.7 The radioactivity of a sample is R₁ at a time T₁ and R₂ at a time T₂. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time (T₂–T₁) is proportional to:

(A) R₁T₁−R₂T₂​

(B) R₁−R₂​

(C) $\frac{R_1-R_2}{T}$

(D)  $(R1-R2)\cdot \frac{T}{0.693}$

Ans: (D) 

Solution:

• Radioactivity R is given by: R = λN
  where:
• • λ is the decay constant
  • N is the number of undecayed nuclei.
• If radioactivity changes from R₁ to R₂​, then the number of atoms that have disintegrated is proportional to: $N_1-N_2=\frac{R_1-R_2}{\lambda }$

But: $\lambda =\frac{0.693}{T}\Rightarrow \frac{1}{\lambda }=\frac{T}{0.693}$

Thus:

$N_1-N_2\propto \frac{R_1-R_2}{\lambda }R1-R2=(R1-R2)\cdot \frac{t}{0.693}$

Ans: (D)  $(R_1-R_2)\cdot \frac{T}{0.693}$

Q. 8 An element X decays first by positron emission, and then two α-particles are emitted in successive radioactive decay.  If the product nucleus has a mass number of 229 and atomic number 89, the mass number and atomic number of element X are:

(A) 273, 93
(B) 237, 94
(C) 238, 93
(D) 273, 92

Ans: (B) 237, 94

Solution: Let:

Final product: A=229, Z=89

Trace back from the final nucleus to the original nucleus X.

Step 1:( α-decay)

Each α-particle reduces:

Mass number (A) by 4
Atomic number (Z) by 2

So, after reversing 2 α-decays:

New mass number = 229 + 4×2 = 237

New atomic number = 89 + 2×2 = 93

Step 2: (β⁺ decay)

A positron emission decreases atomic number by 1 (proton → neutron), so reversing it increases atomic number by 1.

Final atomic number = 93 + 1 = 94

Mass number remains unchanged = 237

Ans: 237, 94

Q.9  Which of the following statements is incorrect?

(A) A Positron has the same mass as that of an electron
(B) Positron has the same charge as that of a proton
(C) Positron has the same e/m ratio as α-particles
(D) Emission of a positron increases the number of protons in the daughter nuclide

Ans: Statement C and Statement D are incorrect.

Solution 

Statement (A). A positron has the same mass as an electron. This statement is correct. 

Statement (B). A positron has a charge of +e. A proton has a charge of +e. Therefore, a positron has the same charge as a proton. This statement is correct.

Statement (C) The charge-to-mass ratio for a positron is $\frac{e}{m_e}$

The charge-to-mass ratio for an alpha particle is  $\frac{2e}{4m_p}=\frac{e}{2m_p}$

Since $m_e\ne 2m_p$,  the e/m ratios are not the same. 

This statement is incorrect. 

Statement (D) Positron emission involves a proton converting to a neutron:  

${}_1^{1}p{\to }_0^{1}n{+}_{+1}^{0}e$

This process decreases the number of protons by 1. 

Therefore, positron emission results in a decrease, not an increase, in the number of protons. This statement is incorrect

Q.10 Radium occurs only in uranium ores, typically with an observed Radium/Uranium ratio of 1 mg / 3 kg. Uranium ores normally contain only about 200 ppm of uranium.
How many kilograms of uranium ore must be processed to obtain 1 mg of radium?

 (A) 1700 kg ore
(B) 15000 kg ore
(C) 6.0×1086.0 \times 10^86.0×108 kg ore
(D) None of the above

Answer: (B) 15000 kg ore

Solution:

Step 1: Understand the data

• Radium/Uranium = 1 mg Ra per 3 kg U
• Uranium content = 200 ppm = 200 mg U / 1 kg ore = 0.2 g U / kg ore

Step 2: How much uranium is needed for 1 mg radium?

From the ratio:

1 mg Ra⇒3 kg of U

So, we need 3 kg of uranium to get 1 mg radium.

Step 3: Ore is needed to get 3 kg uranium

Since each 1 kg ore has 0.2 g = 0.0002 kg U, then:

Ore required=   $\frac{3kgU}{0.0002kgU/kgore}=15000kgore$