Atomic Physics JEE Main PYQs PDF, 2020–2026

Atomic Physics questions per year

Topic-wise weightage by year

How many Atomic Physics questions each topic had in every JEE Main year (all shifts), counting every question asked, including ones not in the PDF. Darker cells mean more questions.

Swipe the table sideways for older years.

Newest questions (2026)

All 11 JEE Main 2026 Atomic Physics questions in the free PDF, latest shift first.

1. JEE Main 2026 · 6 Apr · Shift 1 · Bohr's Model

   The ratio of momentum of the photons of the $1^{{\text{st}}}$ and $2^{{\text{nd}}}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$. The possible values of $\alpha$ and $\beta$ are:-

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2. JEE Main 2026 · 5 Apr · Shift 1 · Bohr's Model

   In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit $(f)$. The ratio of the radius of the orbits in given by $r_{i}:r_{f}=16:4$. The wavelength of photon emitted due to this transition is $\_\_\_\_$ nm.

   (Given Rydberg constant $=1.0973\times {10}^{7}/{\mathrm{m}}$ )

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3. JEE Main 2026 · 4 Apr · Shift 2 · Thomson and Rutherford Models

   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 :

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4. JEE Main 2026 · 2 Apr · Shift 1 · Bohr's Model

   Angular momentum of an electron in a hydrogen atom is $\frac{3h}{\pi}$, then the energy of the electron is _____ eV.

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5. JEE Main 2026 · 2 Apr · Shift 2 · Bohr's Model

   Using Bohr’s model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2^nd and 4^th orbits of hydrogen atom ________.

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6. JEE Main 2026 · 24 Jan · Shift 1 · Bohr's Model

   Two electrons are moving in orbits of two hydrogen like atoms with speeds $3\times {10}^{5}{\ \mathrm{m}}/{\mathrm{s}}$ and $2.5\times {10}^{5}{\ \mathrm{m}}/{\mathrm{s}}$ respectively. If the radii of these orbits are nearly same then the possible order of energy states are $\_\_\_\_$ respectively.

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7. JEE Main 2026 · 23 Jan · Shift 1 · Bohr's Model

   In hydrogen atom spectrum, ( $R\rightarrow$ Rydberg's constant)

   A. the maximum wavelength of the radiation of Lyman series is $\frac{4}{3R}$

   B. the Balmer series lies in the visible region of the spectrum

   C. the minimum wavelength of the radiation of Paschen series is $\frac{9}{R}$

   D. the minimum wavelength of Lyman series is $\frac{5}{4R}$

   Choose the correct answer from the options given below :

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8. JEE Main 2026 · 22 Jan · Shift 2 · Bohr's Model

   The smallest wavelength of Lyman series is 91 nm . The difference between the largest wavelengths of Paschen and Balmer series is nearly $\_\_\_\_$ nm.

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9. JEE Main 2026 · 22 Jan · Shift 1 · Thomson and Rutherford Models

   $7.9{\mathrm{MeV}}\alpha$-particle scatters from a target material of atomic number 79 . From the given data the estimated diameter of nuclei of the target material is (approximately) $\_\_\_\_$ m.

   $\left[\frac{1}{4\pi\epsilon_{{{\mathrm{o}}}}}=9\times {10}^{9}{\mathrm{Nm}}^{2}/{\mathrm{C}}^{2}\text{ and electron charge}=1.6\times {10}^{{-19}}{\mathrm{C}}\right]$

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10. JEE Main 2026 · 21 Jan · Shift 1 · Thomson and Rutherford Models

    If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is $\_\_\_\_$ m. (Atomic number of gold $=79$ and $\frac{1}{4\pi\epsilon_{{{\mathrm{o}}}}}=9\times {10}^{9}$ in SI units)

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11. JEE Main 2026 · 21 Jan · Shift 2 · Bohr's Model

    The energy of an electron in an orbit of the Bohr's atom is $-0.04E_{0}$ eV where $E_{0}$ is the ground state energy. If $L$ is the angular momentum of the electron in this orbit and $h$ is the Planck's constant, then

    $\frac{2\pi L}{h}$ is ________ :

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