PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 12 · JEE Main

Atoms

Chapter-wise practice and revision for JEE Main: concise notes on Rutherford scattering, the Bohr model of hydrogen-like atoms, and the hydrogen spectrum, concept figures, and 48 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

48 selected questions3 topics2 concept figuresPreview · first 3 of 48 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 109 previous-year JEE questions in the PiTheory bank

Spectrum59
Bohr model44
Rutherford6

Contents

  1. 1Rutherford's nuclear modelQ1–Q5
  2. 2Bohr model of hydrogen-like atomsQ6–Q29
  3. 3Hydrogen spectrum and transitionsQ30–Q48
  4. ✓Last-day checklist
01

Topic 1 · Q1–Q5

Rutherford's nuclear model

6 of the 109 JEE PYQs in the bank test this as their main idea

Quick notes

$$\text{closest approach: }\tfrac12mv^2=\frac{k(2e)(Ze)}{d}\ \Rightarrow\ d=\frac{4kZe^2}{mv^2}$$

Most alpha particles pass straight through; a few bounce back. So the positive charge and nearly all the mass sit in a tiny nucleus ($\sim10^{-15}$ m) inside an atom $\sim10^{-10}$ m across.

  • Small impact parameter (nearly head-on) gives large-angle scattering.
  • Number scattered falls steeply with angle ($\propto1/\sin^4(\theta/2)$).
  • Classical flaw: an orbiting electron radiates and should spiral into the nucleus.
Trap. The alpha particle's charge is $2e$.

Practice: first 3 of 5 questions

Q1AIEEE 2006Medium

An alpha nucleus of energy $\dfrac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

  1. A$\dfrac{1}{v}$
  2. B$\dfrac{1}{Ze}$
  3. C$v^2$
  4. D$\dfrac{1}{m}$
Answer and solution

Answer: Option D: $\dfrac{1}{m}$

Concept. At closest approach in a head-on collision all kinetic energy has become electric potential energy: $\tfrac12mv^2=\dfrac{k(2e)(Ze)}{d}$.

Step 1. Distance. $d=\dfrac{4kZe^2}{mv^2}\propto\dfrac{1}{m}$, $\dfrac1{v^2}$ and $Z$.

Answer. $\dfrac1m$ (Option D).

Point to note. $d\propto1/mv^2$.

Q2JEE Main 2024 (8 Apr, Shift 1)Medium

In an alpha particle scattering experiment, the distance of closest approach for the $\alpha$ particle is $4.5\times10^{-14}\ \text{m}$. If the target nucleus has atomic number 80, then the maximum velocity of the $\alpha$-particle is ______ $\times10^5\ \text{m/s}$ approximately. $\left(\dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{SI unit}, \text{mass of}\ \alpha\ \text{particle} = 6.72\times10^{-27}\ \text{kg}\right)$

Numerical answer type

Answer and solution

Answer: 156

Concept. Same energy balance: $\tfrac12mv^2=\dfrac{k(2e)(Ze)}{d}$.

Step 1. Energy. $\dfrac{9\times10^9\times2\times80\times(1.6\times10^{-19})^2}{4.5\times10^{-14}}=8.19\times10^{-13}$ J.

Step 2. Speed. $v=\sqrt{\dfrac{2\times8.19\times10^{-13}}{6.72\times10^{-27}}}=1.56\times10^7$ m/s.

Answer. $156$.

Point to note. $K=k(2e)(Ze)/d$.

Q3JEE Main 2026 (4 Apr, Shift 2)Easy

In Rutherford's alpha-particle scattering experiment, only a few alpha particles rebound back because

A. The size of gold nucleus is very small as compared to the size of gold atom.

B. Alpha particle and gold nucleus have equal charge.

C. The impact parameter is minimum for a few alpha particles.

D. A few alpha particles have very high kinetic energy.

E. Only a few alpha particles undergo head-on collision with the nuclei.

Choose the correct answer from the options given below :

  1. AA, B Only
  2. BB, E Only
  3. CC, D Only
  4. DA, C, E Only
Answer and solution

Answer: Option D: A, C, E Only

Concept. Large-angle scattering needs a nearly head-on approach (small impact parameter) to a tiny, heavy nucleus.

Step 1. Check. A: true, the nucleus is tiny. B: false; charges are $2e$ and $79e$. C: true. D: all particles had about the same energy, false. E: true.

Answer. A, C, E only (Option D).

Point to note. Small impact parameter.

45 more questions and 2 more topics in the full chapter

  1. 2Bohr model of hydrogen-like atoms24 questions with solutions
  2. 3Hydrogen spectrum and transitions19 questions with solutions
  3. ✓Last-day checklist

The complete Atoms chapter is in the Class 12 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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